Free math worksheets

- Think of the x-coordinate on the unit circle. - Identify where that coordinate reaches its maximum value during the stated half-open interval.

1. On one full unit-circle rotation, cosine equals \(1\) at \(x=0\). The endpoint \(2\pi\) is not included.

\(x=0\)

- Recall the unit-circle angle in the stated interval where the vertical coordinate is zero. - Check that tangent is defined at the angle you choose.

1. Tangent is zero when sine is zero and cosine is nonzero. In the stated interval, this occurs at \(x=0\).

\(x=0\)

- On the unit circle, identify the points whose y-coordinate is zero. - Check the interval carefully: the right endpoint \(2\pi\) is not included.

1. Sine is the y-coordinate on the unit circle, so it is zero where the circle meets the x-axis. 2. On \([0,2\pi)\), those angles are \(x=0\) and \(x=\pi\).

\(x=0\) or \(x=\pi\)

- Which coordinate on the unit circle represents sine, and which represents cosine? - In which quadrants is sine positive or negative? - Which unit-circle points lie on the coordinate axes?

1. For a), sine is the y-coordinate on the unit circle. The y-coordinate is \(0.5\) at \(30^\circ\) and \(150^\circ\). 2. For b), cosine is the x-coordinate. The x-coordinate is \(0\) at the top and bottom of the unit circle, so \(\alpha=90^\circ\) or \(270^\circ\). 3. For c), the y-coordinate is \(-1\) only at the bottom of the unit circle, so \(\alpha=270^\circ\).

a) \(\alpha=30^\circ\) or \(\alpha=150^\circ\) b) \(\alpha=90^\circ\) or \(\alpha=270^\circ\) c) \(\alpha=270^\circ\)

- Identify the quadrants in which sine or cosine has the same sign. - Use reflection symmetry across the coordinate axes. - Angles with equal sine values have equal y-coordinates; angles with equal cosine values have equal x-coordinates.

1. For a), the angles with the same positive sine value are \(80^\circ\) and \(180^\circ-80^\circ=100^\circ\). 2. For b), the angles with the same positive cosine value are \(10^\circ\) and \(360^\circ-10^\circ=350^\circ\). 3. For c), \(200^\circ\) has reference angle \(20^\circ\) and a negative sine value. The other angle with the same sine is \(360^\circ-20^\circ=340^\circ\). 4. For d), \(300^\circ\) has reference angle \(60^\circ\) and positive cosine. The other angle is \(60^\circ\).

a) \(80^\circ, 100^\circ\) b) \(10^\circ, 350^\circ\) c) \(200^\circ, 340^\circ\) d) \(60^\circ, 300^\circ\)

- Use inverse sine to find the Quadrant I solution. - Where else in one rotation does sine have the same positive value? - The two solutions in \([0, \pi]\) are supplementary.

1. The first solution is \(x_1=\sin^{-1}(0.3)\approx0.3047\). 2. Sine has the same positive value in Quadrants I and II, so the second solution is \(x_2=\pi-x_1\approx2.8369\). 3. Rounded to the nearest hundredth, \(x_1\approx0.30\) and \(x_2\approx2.84\).

\(x\approx0.30\) or \(x\approx2.84\)

- Use inverse cosine to find one solution. - Reflect the corresponding unit-circle point across the x-axis. - Make sure the calculator is in radian mode. - Keep full precision until the final rounding step.

1. The first solution is \(x_1=\cos^{-1}(-0.65)\approx2.2784\). 2. Cosine has the same value at the reflected angle \(x_2=2\pi-x_1\approx4.0048\). 3. Rounded to the nearest hundredth, the solutions are \(2.28\) and \(4.00\).

\(x\approx2.28\) or \(x\approx4.00\)

- Isolate the sine expression first. - In which quadrants is sine negative? - Use the reference angle for a sine value of \(\frac{1}{2}\).

1. Subtract \(12\): \(6\sin(x)=-3\). 2. Divide by \(6\): \(\sin(x)=-\frac{1}{2}\). 3. The reference angle is \(\frac{\pi}{6}\). Sine is negative in Quadrants III and IV, so \(x=\pi+\frac{\pi}{6}=\frac{7\pi}{6}\) or \(x=2\pi-\frac{\pi}{6}=\frac{11\pi}{6}\).

\(x=\frac{7\pi}{6}\) or \(x=\frac{11\pi}{6}\)

- Recall the special-angle cosine value \(\frac{\sqrt{2}}{2}\). - Determine the quadrants in which cosine is negative. - Place the reference angle in those quadrants.

1. The reference angle for a cosine magnitude of \(\frac{\sqrt{2}}{2}\) is \(\frac{\pi}{4}\). 2. Cosine is negative in Quadrants II and III. 3. Therefore, \(x=\pi-\frac{\pi}{4}=\frac{3\pi}{4}\) or \(x=\pi+\frac{\pi}{4}=\frac{5\pi}{4}\).

\(x=\frac{3\pi}{4}\) or \(x=\frac{5\pi}{4}\)

- First isolate \(\cos(x)\). - Determine where cosine is negative. - Find the reference angle with cosine magnitude \(\frac{\sqrt{3}}{2}\). - Place the reference angle in the appropriate quadrants.

1. Isolate cosine: \(2\cos(x)=-\sqrt{3}\), so \(\cos(x)=-\frac{\sqrt{3}}{2}\). 2. The reference angle for a cosine magnitude of \(\frac{\sqrt{3}}{2}\) is \(\frac{\pi}{6}\). 3. Cosine is negative in Quadrants II and III. 4. Therefore, \(x=\pi-\frac{\pi}{6}=\frac{5\pi}{6}\) or \(x=\pi+\frac{\pi}{6}=\frac{7\pi}{6}\).

\(x=\frac{5\pi}{6}\) or \(x=\frac{7\pi}{6}\)

- In which quadrants is sine negative? - Use the Quadrant I reference angle for a sine magnitude of \(\frac{1}{2}\). - Transfer that reference angle to the appropriate quadrants.

1. The reference angle for \(|\sin(x)|=\frac{1}{2}\) is \(\frac{\pi}{6}\). 2. Sine is negative in Quadrants III and IV. 3. Therefore, \(x=\pi+\frac{\pi}{6}=\frac{7\pi}{6}\) or \(x=2\pi-\frac{\pi}{6}=\frac{11\pi}{6}\).

\(\left\{\frac{7\pi}{6}, \frac{11\pi}{6}\right\}\)

- Think about the sign of sine in each quadrant. - A non-extreme sine value generally occurs twice in one full rotation.

1. The sine value \(\frac{1}{2}\) is positive and is not an extreme value. 2. Sine is positive in Quadrants I and II, giving one solution in each quadrant on one full rotation. 3. Therefore, there are two solutions.

There are \(2\) solutions, one in Quadrant I and one in Quadrant II.

- Interpret sine as the vertical coordinate on the unit circle. - Look at which two quadrants contain the marked chord endpoints. - Use the familiar reference angle associated with a vertical coordinate of one-half.

1. The marked height is \(\frac{1}{2}\) of the unit radius, so the chord represents \(\sin(x)=\frac{1}{2}\). 2. The two marked endpoints lie in Quadrants I and II. 3. The reference angle for sine value \(\frac{1}{2}\) is \(\frac{\pi}{6}\), so \(x=\frac{\pi}{6}\) or \(x=\frac{5\pi}{6}\). 4. The horizontal chord meets the circle at exactly those two endpoints, so there are exactly two solutions in one rotation.

a) \(\sin(x)=\frac{1}{2}\) b) \(x=\frac{\pi}{6}\) or \(x=\frac{5\pi}{6}\); there are two solutions because the marked horizontal chord meets the unit circle at two points.

- Identify the familiar reference angle whose tangent value is \(\sqrt{3}\). - Determine the two quadrants where tangent is positive. - Compare tangent's period with the period of sine and cosine.

1. The reference angle is \(\frac{\pi}{3}\) because \(\tan\left(\frac{\pi}{3}\right)=\sqrt{3}\). 2. Tangent is positive in Quadrants I and III, so the solutions are \(x=\frac{\pi}{3}\) and \(x=\frac{4\pi}{3}\). 3. Tangent has period \(\pi\), so corresponding tangent values repeat after \(\pi\) radians; indeed, \(\frac{4\pi}{3}-\frac{\pi}{3}=\pi\).

\(x=\frac{\pi}{3}\) or \(x=\frac{4\pi}{3}\). The solutions differ by \(\pi\) because tangent has period \(\pi\).

- Find the reference angle associated with the given sine value. - Identify both quadrants where sine has the required sign before introducing periodicity. - Each solution in one rotation generates a full family by adding complete sine periods.

1. The reference angle is \(\frac{\pi}{3}\). 2. Sine is negative in Quadrants III and IV, so one full rotation gives \(x=\frac{4\pi}{3}\) and \(x=\frac{5\pi}{3}\). 3. Sine has period \(2\pi\), so add \(2\pi k\) to each family. 4. The complete solution set is \(x=\frac{4\pi}{3}+2\pi k\) or \(x=\frac{5\pi}{3}+2\pi k\), where \(k\in\mathbb Z\).

\(x=\frac{4\pi}{3}+2\pi k\) or \(x=\frac{5\pi}{3}+2\pi k\), where \(k\in\mathbb Z\)

- Translate the equation into a relationship between the x- and y-coordinates on the unit circle. - In which quadrants do sine and cosine have opposite signs? - For which reference angle do sine and cosine have equal absolute values? - Consider where the line \(y=-x\) intersects the unit circle.

1. On the unit circle, \(\sin(\alpha)=-\cos(\alpha)\) means the coordinates satisfy \(y=-x\). 2. Sine and cosine must have opposite signs, so the points lie in Quadrants II and IV. 3. Their absolute values must be equal, which gives a reference angle of \(45^\circ\). 4. In Quadrant II, \(\alpha=180^\circ-45^\circ=135^\circ\). 5. In Quadrant IV, \(\alpha=360^\circ-45^\circ=315^\circ\).

\(\alpha=135^\circ\) or \(\alpha=315^\circ\)

- Use the Pythagorean identity. - Remember both signs when taking a square root. - Determine where sine is positive. - Use the quadrant to choose the cosine sign.

1. From the Pythagorean identity, \(\cos^2(\alpha)=1-(0.6)^2=0.64\), so \(\cos(\alpha)=0.8\) or \(\cos(\alpha)=-0.8\). 2. The first angle is \(\alpha_1=\sin^{-1}(0.6)\approx36.9^\circ\). 3. Sine is also positive in Quadrant II, so \(\alpha_2=180^\circ-36.9^\circ\approx143.1^\circ\). 4. In Quadrant I, cosine is positive, so \(\cos(\alpha_1)=0.8\). In Quadrant II, cosine is negative, so \(\cos(\alpha_2)=-0.8\).

a) \(\cos(\alpha)=0.8\) or \(\cos(\alpha)=-0.8\) b) \(\alpha\approx36.9^\circ\) or \(\alpha\approx143.1^\circ\) c) For \(\alpha\approx36.9^\circ\), \(\cos(\alpha)=0.8\); for \(\alpha\approx143.1^\circ\), \(\cos(\alpha)=-0.8\)

- Isolate the trigonometric function first. - Use an inverse trigonometric function to find a reference angle. - For part c), angles with the same sine value have the same y-coordinate on the unit circle.

1. For a), \(\cos(\alpha)=0.75\). The reference angle is \(\cos^{-1}(0.75) \approx 41.41^\circ\). Cosine is positive in Quadrants I and IV, so \(\alpha \approx 41.41^\circ\) or \(360^\circ-41.41^\circ=318.59^\circ\). 2. For b), \(\sin(\alpha)=-0.4\). The reference angle is \(\sin^{-1}(0.4) \approx 23.58^\circ\). Sine is negative in Quadrants III and IV, so \(\alpha \approx 180^\circ+23.58^\circ=203.58^\circ\) or \(360^\circ-23.58^\circ=336.42^\circ\). 3. For c), one solution is \(40^\circ\). The other angle with the same sine value is its supplement: \(180^\circ-40^\circ=140^\circ\).

a) \(\alpha \approx 41.41^\circ\) or \(\alpha \approx 318.59^\circ\) b) \(\alpha \approx 203.58^\circ\) or \(\alpha \approx 336.42^\circ\) c) \(\alpha=40^\circ\) or \(\alpha=140^\circ\)

- Use the quadrant to choose the correct angle from the inverse-sine result. - Use the Pythagorean identity to find cosine and choose its sign from the quadrant. - Equal cosine values correspond to unit-circle points reflected across the x-axis.

1. Since \(\alpha\) is in Quadrant II, \(\alpha=180^\circ-\sin^{-1}(0.75)\approx131.41^\circ\). 2. Using the Pythagorean identity and the Quadrant II sign, \(\cos(\alpha)=-\sqrt{1-(0.75)^2}\approx-0.6614\). 3. An angle with the same cosine is the reflection across the x-axis: \(\beta=360^\circ-131.41^\circ\approx228.59^\circ\). 4. Since \(180^\circ < 228.59^\circ < 270^\circ\), \(\beta\) lies in Quadrant III.

a) \(\alpha\approx131.41^\circ\) b) \(\cos(\alpha)\approx-0.6614\) c) \(\beta\approx228.59^\circ\), in Quadrant III

- When is a product of two factors equal to zero? - When solving \(u^2=a\), include both the positive and negative square roots. - Use the unit circle to locate every angle in the interval.

1. For a), use the zero-product property. Either \(\sin(\alpha)-1=0\) or \(\cos(\alpha)+0.5=0\). 2. The equation \(\sin(\alpha)=1\) gives \(\alpha=90^\circ\). 3. The equation \(\cos(\alpha)=-0.5\) gives \(\alpha=120^\circ\) or \(240^\circ\). 4. Therefore, the solution set for a) is \(\{90^\circ,120^\circ,240^\circ\}\). 5. For b), \(\cos^2(\alpha)=0.25\) gives \(\cos(\alpha)=0.5\) or \(\cos(\alpha)=-0.5\). 6. The positive value occurs at \(60^\circ\) and \(300^\circ\); the negative value occurs at \(120^\circ\) and \(240^\circ\). 7. Therefore, the solution set for b) is \(\{60^\circ,120^\circ,240^\circ,300^\circ\}\).

a) \(\{90^\circ,120^\circ,240^\circ\}\) b) \(\{60^\circ,120^\circ,240^\circ,300^\circ\}\)

- What does it mean geometrically when a unit-circle point has equal x- and y-coordinates? - You may rewrite the first equation using another trigonometric function when cosine is nonzero. - Factor the second equation before solving.

1. For a), \(\sin(\alpha)=\cos(\alpha)\) means the unit-circle coordinates satisfy \(y=x\). 2. The line \(y=x\) intersects the unit circle at angles \(45^\circ\) and \(225^\circ\). 3. For b), factor: \(\cos(\alpha)(\cos(\alpha)+1)=0\). 4. If \(\cos(\alpha)=0\), then \(\alpha=90^\circ\) or \(270^\circ\). 5. If \(\cos(\alpha)=-1\), then \(\alpha=180^\circ\). 6. Therefore, the solutions to b) are \(90^\circ\), \(180^\circ\), and \(270^\circ\).

a) \(\{45^\circ,225^\circ\}\) b) \(\{90^\circ,180^\circ,270^\circ\}\)

- Recall the exact sine and cosine values at \(45^\circ\). - Use a reference angle and the specified quadrant. - What is the range of the sine function?

1. Since \(\cos(45^\circ)=\sin(45^\circ)=\frac{\sqrt{2}}{2}\), solve \(\sin(\alpha)=\sin(45^\circ)\). The solutions are \(45^\circ\) and \(135^\circ\). 2. The reference angle for \(|\cos(\alpha)|=0.5\) is \(60^\circ\). In Quadrant III, \(\alpha=180^\circ+60^\circ=240^\circ\). 3. Sine is the y-coordinate on the unit circle, so its range is \([-1, 1]\). Therefore, it cannot equal \(1.2\).

a) \(45^\circ, 135^\circ\) b) \(240^\circ\) c) No solution, because the range of sine is \([-1, 1]\).

- How many solutions should a nonextreme sine value have in one full rotation? - Identify the quadrants in which sine has the required sign. - Use an inverse sine to find a reference angle, then locate the second angle by symmetry.

1. For a), the reference angle is \(\sin^{-1}(0.35)\approx20.5^\circ\). Sine is positive in Quadrants I and II, so the solutions are \(20.5^\circ\) and \(180^\circ-20.5^\circ\approx159.5^\circ\). 2. For b), the reference angle is \(\sin^{-1}(0.82)\approx55.1^\circ\). Sine is negative in Quadrants III and IV, so the solutions are \(180^\circ+55.1^\circ\approx235.1^\circ\) and \(360^\circ-55.1^\circ\approx304.9^\circ\).

a) \(\alpha\approx20.5^\circ\) or \(\alpha\approx159.5^\circ\) b) \(\alpha\approx235.1^\circ\) or \(\alpha\approx304.9^\circ\)

- Cosine is the x-coordinate on the unit circle. - Which two points in one full rotation can have the same x-coordinate? - Recall the exact cosine values for the special angles \(30^\circ\), \(45^\circ\), and \(60^\circ\). - Use symmetry across the x-axis to find the second angle.

1. For a), \(\cos^{-1}(-0.4)\approx113.6^\circ\). Cosine is negative in Quadrants II and III, and the second solution is \(360^\circ-113.6^\circ\approx246.4^\circ\). 2. For b), \(\frac{\sqrt{3}}{2}\) is the cosine of \(30^\circ\). Cosine is positive in Quadrants I and IV, so the solutions are \(30^\circ\) and \(360^\circ-30^\circ=330^\circ\).

a) \(\alpha\approx113.6^\circ\) or \(\alpha\approx246.4^\circ\) b) \(\alpha=30^\circ\) or \(\alpha=330^\circ\)

- Use the x- or y-coordinate on the unit circle, depending on the function. - Most nonextreme sine and cosine values occur twice in one rotation. - Use a reference angle and the signs in each quadrant. - Move a negative inverse-trigonometric result into \([0, 2\pi)\) using the period.

1. For a), the reference angle is \(\frac{\pi}{6}\). Sine is positive in Quadrants I and II, so \(x=\frac{\pi}{6}\approx0.52\) or \(x=\frac{5\pi}{6}\approx2.62\). 2. For b), cosine is zero at the top and bottom of the unit circle, so \(x=\frac{\pi}{2}\approx1.57\) or \(x=\frac{3\pi}{2}\approx4.71\). 3. For c), the reference angle is \(\sin^{-1}(0.6)\approx0.64\). Sine is negative in Quadrants III and IV, so \(x\approx\pi+0.64\approx3.79\) or \(x\approx2\pi-0.64\approx5.64\).

a) \(x=\frac{\pi}{6}\) or \(x=\frac{5\pi}{6}\) b) \(x=\frac{\pi}{2}\) or \(x=\frac{3\pi}{2}\) c) \(x\approx3.79\) or \(x\approx5.64\)

- Visualize intersections of the graph with a horizontal line. - An extreme value such as \(1\) may occur only once in one period. - Use unit-circle symmetry to find a second solution when appropriate. - Use radian mode for calculator values.

1. For a), the reference angle is \(\frac{\pi}{3}\), and cosine is negative in Quadrants II and III. Thus \(x=\frac{2\pi}{3}\) or \(x=\frac{4\pi}{3}\). 2. For b), sine reaches its maximum value \(1\) only at \(x=\frac{\pi}{2}\) in the interval. 3. For c), \(x_1=\cos^{-1}(0.25)\approx1.318\). The second solution is \(x_2=2\pi-x_1\approx4.965\).

a) \(x=\frac{2\pi}{3}\) or \(x=\frac{4\pi}{3}\) b) \(x=\frac{\pi}{2}\) c) \(x\approx1.318\) or \(x\approx4.965\)

- Isolate the entire sine expression first. - A substitution for the shifted input can simplify the equation. - Track how the substitution changes the interval. - Use the unit circle to locate where sine equals \(-\frac{\sqrt{2}}{2}\).

1. Isolate the sine expression: \(\sin\left(x+\frac{\pi}{3}\right)=-\frac{\sqrt{2}}{2}\). 2. Let \(u=x+\frac{\pi}{3}\). Then \(u \in \left[\frac{\pi}{3}, \frac{7\pi}{3}\right]\). 3. In this interval, \(\sin(u)=-\frac{\sqrt{2}}{2}\) at \(u=\frac{5\pi}{4}\) and \(u=\frac{7\pi}{4}\). 4. Subtract \(\frac{\pi}{3}\): \(x=\frac{5\pi}{4}-\frac{\pi}{3}=\frac{11\pi}{12}\) or \(x=\frac{7\pi}{4}-\frac{\pi}{3}=\frac{17\pi}{12}\).

\(x \in \left\{\frac{11\pi}{12}, \frac{17\pi}{12}\right\}\)

- Set the two function expressions equal. - Which line contains points whose y-coordinate is the opposite of their x-coordinate? - Find the solutions in one rotation, then use periodicity for the second rotation.

1. Set the functions equal: \(\sin(\alpha)=-\cos(\alpha)\). 2. On the unit circle, this means the point satisfies \(y=-x\). 3. During the first rotation, the line \(y=-x\) meets the unit circle at \(135^\circ\) and \(315^\circ\). 4. The functions repeat every \(360^\circ\), so add \(360^\circ\) to obtain \(495^\circ\) and \(675^\circ\). 5. Neither endpoint \(0^\circ\) nor \(720^\circ\) satisfies the equation.

\(\alpha\in\{135^\circ, 315^\circ, 495^\circ, 675^\circ\}\)

- Determine the signs of sine and cosine in each quadrant. - Cosine is the x-coordinate. - Use inverse cosine, then verify the quadrant.

1. A negative cosine places the angle in Quadrant II or III. A positive sine places it in Quadrant I or II. Therefore, the angle must lie in Quadrant II. 2. Using inverse cosine, \(\alpha=\cos^{-1}(-0.5736)\approx125.0^\circ\). 3. This angle lies in Quadrant II and satisfies both sign conditions.

a) Quadrant II b) \(\alpha\approx125.0^\circ\)

- Determine how the factor \(2\) changes the period. - First solve the equation over one period. - Generate additional solutions by adding or subtracting the least period. - Check both the negative and positive portions of the interval.

1. Divide by \(2\): \(\sin(2\alpha)=\frac{1}{2}\). 2. In one cycle, \(2\alpha=30^\circ\) or \(2\alpha=150^\circ\). Thus, \(\alpha=15^\circ\) or \(\alpha=75^\circ\). 3. The least period of \(\sin(2\alpha)\) is \(\frac{360^\circ}{2}=180^\circ\). 4. Add integer multiples of \(180^\circ\) to each initial solution and retain the values in the given interval. 5. From \(15^\circ\): \(-345^\circ, -165^\circ, 15^\circ, 195^\circ, 375^\circ, 555^\circ\). 6. From \(75^\circ\): \(-285^\circ, -105^\circ, 75^\circ, 255^\circ, 435^\circ, 615^\circ\).

\(\alpha \in \{-345^\circ, -285^\circ, -165^\circ, -105^\circ, 15^\circ, 75^\circ, 195^\circ, 255^\circ, 375^\circ, 435^\circ, 555^\circ, 615^\circ\}\)

- First solve for the angle inside the cosine function. - Multiplying the input by \(0.5\) doubles the period. - After solving for \(\alpha\), add or subtract the new period. - Keep only values in the stated interval.

1. Let \(u=0.5\alpha\). The equation \(\cos(u)=-0.5\) has base solutions \(u=120^\circ\) and \(u=240^\circ\). 2. Since \(\alpha=2u\), these give \(\alpha=240^\circ\) and \(\alpha=480^\circ\). 3. The function \(\cos(0.5\alpha)\) has period \(\frac{360^\circ}{0.5}=720^\circ\). 4. Subtracting \(720^\circ\) from \(480^\circ\) gives the additional solution \(-240^\circ\). All other period shifts lie outside the interval.

\(\alpha \in \{-240^\circ, 240^\circ, 480^\circ\}\)

- Expect two solutions in a full period for this nonextreme sine value. - Use unit-circle symmetry to find a second solution. - Add or subtract multiples of the period \(2\pi\). - Keep only values in the stated interval.

1. The principal inverse-sine value is \(\sin^{-1}(-0.45)\approx-0.4668\). 2. A second solution in one cycle is \(\pi-(-0.4668)\approx3.6084\). 3. For a), move the negative principal value into \([0, 2\pi]\): \(-0.4668+2\pi\approx5.8164\). Thus the solutions are approximately \(3.61\) and \(5.82\). 4. For b), include the solutions from the adjacent cycle. Subtracting \(2\pi\) from \(3.6084\) gives approximately \(-2.6748\), and the principal value is approximately \(-0.4668\). Together with the two positive solutions, the four rounded values are \(-2.67\), \(-0.47\), \(3.61\), and \(5.82\).

a) \(\{3.61, 5.82\}\) b) \(\{-2.67, -0.47, 3.61, 5.82\}\)

- First isolate \(\cos(x)\). - Use symmetry to find two solutions in one period. - Add and subtract the period \(2\pi\). - Keep only values in the requested interval.

1. Divide by \(5\) to obtain \(\cos(x)=0.6\). 2. The first solution is \(x_1=\cos^{-1}(0.6)\approx0.9273\). The second solution in one period is \(x_2=2\pi-x_1\approx5.3559\). 3. For the larger interval, add and subtract \(2\pi\) from the two base solutions. 4. This gives approximately \(-5.3559\), \(-0.9273\), \(0.9273\), \(5.3559\), \(7.2105\), and \(11.6391\), all of which lie in \([-2\pi, 4\pi]\). 5. Rounding to the nearest hundredth gives the listed solution sets.

a) \(\{0.93, 5.36\}\) b) \(\{-5.36, -0.93, 0.93, 5.36, 7.21, 11.64\}\)

- Isolate the sine expression before using inverse sine. - In part b, determine how the coefficient \(2\) changes the period. - A substitution can reduce part b to a basic cosine equation. - Convert back to the original variable after solving.

1. For part a, divide by \(2.5\): \(\sin(x)=0.6\). The first solution is \(x=\arcsin(0.6)\approx0.64\), and the second is \(x=\pi-\arcsin(0.6)\approx2.50\). 2. For part b, let \(u=2x\), so \(0 \leq u \leq 4\pi\). 3. In this interval, \(\cos(u)=0.5\) at \(u=\frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{11\pi}{3}\). 4. Dividing by \(2\) gives \(x=\frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}\), or approximately \(0.52, 2.62, 3.67, 5.76\).

a) \(x\approx0.64, 2.50\) b) \(x\approx0.52, 2.62, 3.67, 5.76\)

- Determine where sine is negative on the unit circle. - Use inverse sine to find one solution. - Use symmetry to find the other solution in the same cycle. - Sine repeats every \(360^\circ\).

1. The principal inverse-sine value is \(\alpha_1=\sin^{-1}(-0.42)\approx-24.83^\circ\). 2. The other solution in the same cycle is \(\alpha_2=-180^\circ-(-24.83^\circ)\approx-155.17^\circ\). 3. Adding or subtracting \(360^\circ\) from either angle gives values outside the required interval, so these are the only solutions.

\(\alpha\approx-155.17^\circ\) or \(\alpha\approx-24.83^\circ\)

- Part a is a direct function evaluation and does not supply a solution to part b. - In part b, isolate the sine expression before introducing a temporary variable for the input. - Transform the interval together with the substitution, then convert the solutions back to \(x\).

1. For part a, \(y_1=4\sin\left(\frac{\pi}{2}\right)=4\). 2. For part b, solve \(4\sin(0.5x)=2\), so \(\sin(0.5x)=\frac{1}{2}\). 3. Let \(u=0.5x\). Then \(0\le u\le2\pi\), and \(\sin(u)=\frac{1}{2}\) at \(u=\frac{\pi}{6}\) and \(u=\frac{5\pi}{6}\). 4. Multiply by \(2\): \(x=\frac{\pi}{3}\) or \(x=\frac{5\pi}{3}\).

a) \(y_1=4\) b) \(x=\frac{\pi}{3}\) or \(x=\frac{5\pi}{3}\)

- Use the symmetry of cosine on the unit circle or its graph. - Determine how the factor \(2\) changes the period. - Expect multiple solutions in \([-\pi, \pi]\). - Check that every solution lies in the stated interval.

1. Substitute \(x=\frac{\pi}{4}\): \(g\left(\frac{\pi}{4}\right)=\cos\left(2\cdot \frac{\pi}{4}\right)=\cos\left(\frac{\pi}{2}\right)=0\). 2. For part b, solve \(\cos(2x)=-\frac{1}{2}\). Let \(u=2x\), so \(-2\pi \leq u \leq 2\pi\). 3. In this interval, \(\cos(u)=-\frac{1}{2}\) at \(u=-\frac{4\pi}{3}, -\frac{2\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}\). 4. Divide each value by \(2\): \(x=-\frac{2\pi}{3}, -\frac{\pi}{3}, \frac{\pi}{3}, \frac{2\pi}{3}\).

a) \(0\) b) \(x \in \left\{-\frac{2\pi}{3}, -\frac{\pi}{3}, \frac{\pi}{3}, \frac{2\pi}{3}\right\}\)

- Substitute the x-coordinate in part a without using it to infer part b by symmetry. - In part b, divide by the amplitude before solving the sine equation. - Use the fact that sine is one-to-one on the stated interval.

1. For part a, \(y_1=1.2\sin\left(-\frac{\pi}{3}\right)=-0.6\sqrt{3}\). 2. For part b, \(0.6=1.2\sin(x_2)\), so \(\sin(x_2)=\frac{1}{2}\). 3. Sine is one-to-one on \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\), so \(x_2=\frac{\pi}{6}\).

a) \(y_1=-0.6\sqrt{3}\) b) \(x_2=\frac{\pi}{6}\)

- Determine how the factor in the argument changes the number of solutions. - First find where sine equals \(\frac{\sqrt{2}}{2}\) in one cycle. - Track how the substitution changes the interval. - Use sine's symmetry and periodicity to find every solution.

1. Let \(u=2x\). Then \(0 \leq u \leq 4\pi\). 2. In the first cycle, \(\sin(u)=\frac{\sqrt{2}}{2}\) at \(u=\frac{\pi}{4}\) and \(u=\frac{3\pi}{4}\). 3. Add \(2\pi\) to obtain the additional solutions \(u=\frac{9\pi}{4}\) and \(u=\frac{11\pi}{4}\). 4. Divide by \(2\): \(x=\frac{\pi}{8}, \frac{3\pi}{8}, \frac{9\pi}{8}, \frac{11\pi}{8}\).

\(x \in \left\{\frac{\pi}{8}, \frac{3\pi}{8}, \frac{9\pi}{8}, \frac{11\pi}{8}\right\}\)

- Rewrite the sine value as a cosine of a complementary angle. - How many unit-circle points can have the same x-coordinate? - Locate the two quadrants in which cosine is positive.

1. Use the cofunction identity: \(\sin(68^\circ)=\cos(90^\circ-68^\circ)=\cos(22^\circ)\). 2. One solution is \(\alpha=22^\circ\). 3. Cosine has the same value at angles reflected across the x-axis, so the second solution is \(360^\circ-22^\circ=338^\circ\).

\(\alpha=22^\circ\) or \(\alpha=338^\circ\)

- Determine the quadrants in which sine or cosine has the required sign. - Use an inverse trigonometric function to obtain one angle, then use symmetry. - Add or subtract \(360^\circ\) to find solutions in adjacent cycles. - Check every value against the interval endpoints.

1. For a), the principal value is \(\sin^{-1}(-0.65)\approx-40.5^\circ\). A second solution in one cycle is \(180^\circ-(-40.5^\circ)\approx220.5^\circ\). 2. Use the \(360^\circ\) period to obtain the remaining solutions in the interval: \(-40.5^\circ+360^\circ=319.5^\circ\) and \(220.5^\circ-360^\circ=-139.5^\circ\). 3. For b), \(\cos^{-1}(0.22)\approx77.3^\circ\). Reflection across the x-axis gives \(-77.3^\circ\). 4. Adding \(360^\circ\) to \(-77.3^\circ\) gives the additional solution \(282.7^\circ\).

a) \(x \approx -139.5^\circ, -40.5^\circ, 220.5^\circ,\text{ or } 319.5^\circ\) b) \(x \approx -77.3^\circ, 77.3^\circ,\text{ or } 282.7^\circ\)

- When taking the square root, include both positive and negative cases. - Determine how many unit-circle positions have the given absolute sine value. - Check which coterminal angles lie in the specified interval. - Use the repeating sine pattern to verify that no solutions are missing.

1. Taking square roots gives two cases: \(\sin x=\frac{\sqrt{3}}{2}\) or \(\sin x=-\frac{\sqrt{3}}{2}\). 2. In \([-\pi, 2\pi]\), the first equation has solutions \(x=\frac{\pi}{3}\) and \(x=\frac{2\pi}{3}\). 3. In the same interval, the second equation has solutions \(x=-\frac{2\pi}{3}, -\frac{\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}\). 4. Combine the two cases.

\(x \in \left\{-\frac{2\pi}{3}, -\frac{\pi}{3}, \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}\right\}\)

- Identify the expression used as the input of sine. - Sine takes the value \(\frac{1}{2}\) twice in each full cycle. - Determine how the coefficient of \(x\) changes the period. - Check every result against the given interval. - Isolate the angle expression before solving for \(x\).

1. Let \(u=2x-60^\circ\). The solutions of \(\sin(u)=\frac{1}{2}\) are \(u=30^\circ+360^\circ k\) or \(u=150^\circ+360^\circ k\), where \(k\) is an integer. 2. From \(2x-60^\circ=30^\circ+360^\circ k\), obtain \(x=45^\circ+180^\circ k\). The values in the interval are \(45^\circ\) and \(225^\circ\). 3. From \(2x-60^\circ=150^\circ+360^\circ k\), obtain \(x=105^\circ+180^\circ k\). The values in the interval are \(105^\circ\) and \(285^\circ\).

\(x \in \{45^\circ, 105^\circ, 225^\circ, 285^\circ\}\)

- Cosine repeats every \(360^\circ\). - The coefficient \(3\) changes the period and the number of solutions. - Determine the quadrants where cosine is negative with reference angle \(45^\circ\). - Include negative integer values when generating solutions.

1. Let \(u=3\beta\). The general solutions of \(\cos(u)=-\frac{\sqrt{2}}{2}\) are \(u=135^\circ+360^\circ k\) and \(u=-135^\circ+360^\circ k\). 2. Divide by \(3\): \(\beta=45^\circ+120^\circ k\) or \(\beta=-45^\circ+120^\circ k\). 3. The first family gives \(-75^\circ, 45^\circ, 165^\circ\) in the interval. 4. The second family gives \(-165^\circ, -45^\circ, 75^\circ\) in the interval.

\(\beta \in \{-165^\circ, -75^\circ, -45^\circ, 45^\circ, 75^\circ, 165^\circ\}\)

- Use the graph first to count the intersections and estimate their locations roughly. - Use inverse sine to refine the first positive intersection numerically. - Use sine symmetry to obtain the second solution in the same rotation.

1. The graph label shows \(g:y=0.3\), so the represented equation is \(\sin(x)=0.3\). The graph shows two intersections, roughly near \(0.3\) and \(2.8\) radians. 2. The first solution is \(x=\sin^{-1}(0.3)\approx0.3047\). 3. The second solution is \(x=\pi-\sin^{-1}(0.3)\approx2.8369\). 4. To the nearest hundredth, the solutions are \(0.30\) and \(2.84\).

a) \(\sin(x)=0.3\); approximately \(x\approx0.3\) and \(x\approx2.8\) b) \(x\approx0.30\) or \(x\approx2.84\)

- First decide how many times a horizontal line at \(y=0.6\) meets one cosine period. - Then count how many complete periods occur in the given interval. - You do not need inverse cosine to answer the question.

1. Since \(0.6\) is strictly between \(-1\) and \(1\), \(\cos(x)=0.6\) has two solutions during each interval of length \(2\pi\). 2. The interval \([0,4\pi)\) contains two complete periods. 3. Therefore, the equation has \(2\cdot2=4\) solutions.

\(4\) solutions

- The factor \(2\) inside sine changes how many cycles appear on the interval. - Temporarily replace \(2x\) with a new variable and transform the interval with it. - After finding the new-variable solutions, convert back to \(x\).

1. The graph shows four intersections on \([0,2\pi)\). 2. Let \(u=2x\). Then \(u\in[0,4\pi)\) and \(\sin(u)=\frac{1}{2}\). 3. In that interval, \(u=\frac{\pi}{6},\frac{5\pi}{6},\frac{13\pi}{6},\frac{17\pi}{6}\). 4. Divide by \(2\): \(x=\frac{\pi}{12},\frac{5\pi}{12},\frac{13\pi}{12},\frac{17\pi}{12}\).

a) \(4\) solutions b) \(x=\frac{\pi}{12},\frac{5\pi}{12},\frac{13\pi}{12},\frac{17\pi}{12}\)

- Use inverse tangent to locate one solution. - Think about how often the tangent graph repeats, not how often sine or cosine repeats. - A complete all-real answer should generate every repeated solution with an integer parameter.

1. The principal solution is \(x=\tan^{-1}(0.4)\approx0.3805\). 2. Tangent has period \(\pi\), so every solution is obtained by adding an integer multiple of \(\pi\). 3. Therefore, the exact solution family is \(x=\tan^{-1}(0.4)+k\pi\), where \(k\in\mathbb Z\). Using the principal angle rounded to the nearest hundredth gives \(x\approx0.38+k\pi\).

\(x=\tan^{-1}(0.4)+k\pi\), where \(k\in\mathbb Z\); equivalently, using the requested rounded principal angle, \(x\approx0.38+k\pi\).

- Count only the intersections inside the half-open interval, paying attention to the breaks in the tangent graph. - Use inverse tangent to refine one intersection numerically. - Once one tangent solution is known, use the function's period to locate the other one.

1. The graph label gives \(g:y=-0.5\), so the represented equation is \(\tan(x)=-0.5\). The graph shows two intersections, roughly near \(-0.5\) and \(2.7\) radians. 2. The principal solution is \(x=\tan^{-1}(-0.5)\approx-0.4636\), so one solution is \(x\approx-0.46\). 3. Tangent has period \(\pi\), so the next solution is \(-0.4636+\pi\approx2.6779\), giving \(x\approx2.68\). 4. The two solutions differ by \(\pi\) because every tangent value repeats after one tangent period.

a) \(\tan(x)=-0.5\); approximately \(x\approx-0.5\) and \(x\approx2.7\) b) \(x\approx-0.46\) and \(x\approx2.68\) c) They differ by \(\pi\) because tangent has period \(\pi\).

- Inverse cosine gives one principal angle, but an all-real cosine equation needs the symmetric angle family as well. - Use cosine's even symmetry to represent the second family compactly. - After accounting for symmetry, use the period of cosine to generate every repetition.

1. The principal angle is \(\alpha=\cos^{-1}(-0.3)\approx1.87549\). 2. Cosine is even, so the two solution families in one full cycle are represented by \(x=\alpha\) and \(x=-\alpha\). 3. Cosine has period \(2\pi\), so all real solutions are \(x=\pm\alpha+2\pi k\), where \(k\in\mathbb Z\). 4. Rounding \(\alpha\) to the nearest hundredth gives \(x\approx\pm1.88+2\pi k\).

\(x=\pm\alpha+2\pi k\), where \(\alpha=\cos^{-1}(-0.3)\) and \(k\in\mathbb Z\); numerically, \(x\approx\pm1.88+2\pi k\).

- How is the distance from \((x, y)\) to each coordinate axis found? - Translate “the same distance” into an equation involving absolute values. - Which lines contain all points that are equally distant from the axes? - How many times do those lines intersect the unit circle?

1. The distance from \(P\) to the x-axis is \(|\sin(\alpha)|\), and the distance to the y-axis is \(|\cos(\alpha)|\). 2. Equal distances require \(|\sin(\alpha)|=|\cos(\alpha)|\), or equivalently \(\sin^2(\alpha)=\cos^2(\alpha)\). 3. This occurs when \(\sin(\alpha)=\cos(\alpha)\) or \(\sin(\alpha)=-\cos(\alpha)\). 4. The first equation gives \(45^\circ\) and \(225^\circ\). The second gives \(135^\circ\) and \(315^\circ\). 5. Geometrically, these are the four intersections of the unit circle with the lines \(y=x\) and \(y=-x\).

\(\alpha\in\{45^\circ, 135^\circ, 225^\circ, 315^\circ\}\)

- Think of \(y=k\) as a horizontal line across one complete sine cycle. - Compare what happens for values inside the range, outside the range, and at the two extreme values. - Pay attention to the half-open interval at the endpoints.

1. If \(-1<k<1\), a horizontal line at \(y=k\) meets one sine cycle twice, including \(k=0\). 2. If \(|k|>1\), there are no solutions. 3. At the extreme values \(k=1\) and \(k=-1\), the line touches the sine graph once in \([0,2\pi)\). 4. Therefore, the required values are \(k=1\) and \(k=-1\).

\(k=1\) or \(k=-1\)