Free math worksheets

- The goal is to leave only cosine, so choose the Pythagorean rearrangement that eliminates sine. - Replace the squared sine term and leave the rest of the equation unchanged.

1. Rewrite \(\sin^2(x)\) as \(1-\cos^2(x)\). 2. The equation becomes \(1-\cos^2(x)+\cos(x)=1\), which involves only \(\cos(x)\).

Use \(\sin^2(x)=1-\cos^2(x)\), giving \(1-\cos^2(x)+\cos(x)=1\).

- Look for a standard double-angle expression on the left side. - Transform the interval when the angle changes from \(x\) to \(2x\). - Convert the resulting angles back to \(x\).

1. Use \(2\sin(x)\cos(x)=\sin(2x)\). 2. The equation becomes \(\sin(2x)=1\). 3. For \(2x\in[0,4\pi)\), \(\sin(2x)=1\) at \(2x=\frac{\pi}{2}\) and \(2x=\frac{5\pi}{2}\). 4. Therefore, \(x=\frac{\pi}{4}\) and \(x=\frac{5\pi}{4}\).

\(x=\frac{\pi}{4},\frac{5\pi}{4}\)

- Eliminate the squared cosine so the equation contains only sine. - Treat the remaining sine value as an algebraic unknown and check which quadratic root is in the range of sine. - A positive non-extreme sine value occurs twice on one full rotation.

1. Use \(\cos^2(x)=1-\sin^2(x)\). The equation becomes \(1-\sin^2(x)=\sin(x)\). 2. Rearrange: \(\sin^2(x)+\sin(x)-1=0\). 3. Let \(s=\sin(x)\). The quadratic gives \(s=\frac{-1\pm\sqrt{5}}{2}\). 4. Only \(s=\frac{\sqrt{5}-1}{2}\) lies in \([-1,1]\). 5. Let \(\alpha=\sin^{-1}\left(\frac{\sqrt{5}-1}{2}\right)\). Since the sine value is positive, \(x=\alpha\) or \(x=\pi-\alpha\).

\(x=\sin^{-1}\left(\frac{\sqrt{5}-1}{2}\right)\) or \(x=\pi-\sin^{-1}\left(\frac{\sqrt{5}-1}{2}\right)\)

- Create the exact product that appears in the sine double-angle identity. - After rewriting, the angle is \(2x\), so its interval is twice the original interval. - Convert every transformed-angle solution back to \(x\).

1. Multiply by \(2\): \(2\sin(x)\cos(x)=\frac{\sqrt{2}}{2}\). 2. Use \(2\sin(x)\cos(x)=\sin(2x)\), giving \(\sin(2x)=\frac{\sqrt{2}}{2}\). 3. For \(2x\in[0,4\pi)\), \(2x=\frac{\pi}{4},\frac{3\pi}{4},\frac{9\pi}{4},\frac{11\pi}{4}\). 4. Divide by \(2\): \(x=\frac{\pi}{8},\frac{3\pi}{8},\frac{9\pi}{8},\frac{11\pi}{8}\).

\(x=\frac{\pi}{8},\frac{3\pi}{8},\frac{9\pi}{8},\frac{11\pi}{8}\)

- Look for an angle-addition pattern before trying to solve the equation term by term. - Rewrite the entire left side as one sine function. - After solving the shifted equation, subtract the shift and check the interval.

1. Recognize the left side as \(\sin\left(x+\frac{\pi}{6}\right)\). 2. The equation becomes \(\sin\left(x+\frac{\pi}{6}\right)=\frac{1}{2}\). 3. The relevant angle values are \(x+\frac{\pi}{6}=\frac{\pi}{6}+2\pi k\) or \(x+\frac{\pi}{6}=\frac{5\pi}{6}+2\pi k\). 4. On \([0,2\pi)\), this gives \(x=0\) and \(x=\frac{2\pi}{3}\).

\(x=0,\frac{2\pi}{3}\)

- Record where each tangent expression is defined before clearing any denominator. - Rewrite \(\tan(2x)\) entirely in terms of \(\tan(x)\). - Do not divide by \(\tan(x)\), because zero tangent values may be solutions.

1. The original equation requires both \(\tan(x)\) and \(\tan(2x)\) to be defined. 2. Let \(t=\tan(x)\). Use \(\tan(2x)=\frac{2t}{1-t^2}\): \(\frac{2t}{1-t^2}=2t\). 3. Where the original left side is defined, \(1-t^2\ne0\). Multiply by that denominator: \(2t=2t(1-t^2)\). 4. Rearranging gives \(2t^3=0\), so \(t=0\). 5. On \([0,2\pi)\), \(\tan(x)=0\) at \(x=0\) and \(x=\pi\). Both make \(\tan(2x)\) defined and satisfy the original equation.

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

- Convert the equation so it involves only one trigonometric function. - After substitution, treat the result as a quadratic in \(\cos(x)\). - Solve both factors and keep every solution in the interval.

1. Replace \(\sin^2(x)\) with \(1-\cos^2(x)\). 2. The equation becomes \(2\left(1-\cos^2(x)\right)+\cos(x)=1\). 3. Rearranging gives \(2\cos^2(x)-\cos(x)-1=0\). 4. Factor: \(\left(2\cos(x)+1\right)\left(\cos(x)-1\right)=0\). 5. Thus, \(\cos(x)=-\frac{1}{2}\) or \(\cos(x)=1\), giving \(x=\frac{2\pi}{3},\frac{4\pi}{3},0\).

\(x=0,\frac{2\pi}{3},\frac{4\pi}{3}\)

- Read the horizontal line's level before writing the equation. - Look for the double-angle identity whose expanded form is a sine-cosine product. - Transform the interval when you replace \(x\) by \(2x\).

1. The graph label gives \(g:y=0.5\), and the four intersections are near \(0.3\), \(1.3\), \(3.4\), and \(4.5\) radians. 2. The represented equation is \(2\sin(x)\cos(x)=\frac{1}{2}\). 3. Use \(2\sin(x)\cos(x)=\sin(2x)\), giving \(\sin(2x)=\frac{1}{2}\). 4. For \(2x\in[0,4\pi)\), \(2x=\frac{\pi}{6},\frac{5\pi}{6},\frac{13\pi}{6},\frac{17\pi}{6}\). 5. Therefore, \(x=\frac{\pi}{12},\frac{5\pi}{12},\frac{13\pi}{12},\frac{17\pi}{12}\).

a) \(g\) has y-value \(0.5\); intersections are approximately at \(0.3\), \(1.3\), \(3.4\), and \(4.5\) radians b) \(2\sin(x)\cos(x)=\frac{1}{2}\), with \(x=\frac{\pi}{12},\frac{5\pi}{12},\frac{13\pi}{12},\frac{17\pi}{12}\)

- Convert the double-angle term so the equation contains only sine. - Treat the result as a quadratic in \(\sin(x)\). - Solve both resulting basic sine equations over the full interval.

1. Use \(\cos(2x)=1-2\sin^2(x)\). 2. The equation becomes \(1-2\sin^2(x)+\sin(x)=0\). 3. Rearrange and factor: \(2\sin^2(x)-\sin(x)-1=0\), so \(\left(2\sin(x)+1\right)\left(\sin(x)-1\right)=0\). 4. Thus, \(\sin(x)=-\frac{1}{2}\) or \(\sin(x)=1\). 5. On \([0,2\pi)\), the solutions are \(x=\frac{7\pi}{6},\frac{11\pi}{6},\frac{\pi}{2}\).

\(x=\frac{\pi}{2},\frac{7\pi}{6},\frac{11\pi}{6}\)

- Look for a double-angle identity matching the difference of the two squares. - After introducing the transformed angle, remember that its interval is twice as long as the original interval. - Use the two marked unit-circle locations in one rotation, then account for the second rotation before converting back to \(x\).

1. Use the double-angle identity \(\cos(2x)=\cos^2(x)-\sin^2(x)\). The equation becomes \(\cos(2x)=-\frac{1}{2}\). 2. Since \(0\le x<2\pi\), the transformed angle satisfies \(0\le2x<4\pi\). 3. In one rotation, the diagram shows the two positions where cosine is \(-\frac{1}{2}\): \(\frac{2\pi}{3}\) and \(\frac{4\pi}{3}\). 4. Over two rotations, \(2x=\frac{2\pi}{3},\frac{4\pi}{3},\frac{8\pi}{3},\frac{10\pi}{3}\). 5. Dividing by \(2\) gives \(x=\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}\).

\(x=\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}\)

- Simplify the quotient using double-angle identities before solving. - Keep track of where the original denominator is zero. - Solve the resulting basic equation only on the stated interval.

1. On the domain of the original expression, \(\frac{1-\cos(2x)}{\sin(2x)}=\tan(x)\). 2. The equation becomes \(\tan(x)=1\). 3. On \(0<x<\pi\), the only solution is \(x=\frac{\pi}{4}\). 4. At \(x=\frac{\pi}{2}\), the original denominator \(\sin(2x)\) is zero, so that value is excluded and was never a valid candidate.

\(x=\frac{\pi}{4}\)

- Check the denominator before simplifying the equation. - Factor \(\sin^2(x)\) using a Pythagorean identity and a difference of squares. - A cancellation is valid only where the canceled factor is nonzero.

1. The original denominator requires \(1-\cos(x)\ne0\), so \(x=0\) and \(x=2\pi\) are excluded. 2. For every remaining \(x\), use \(\sin^2(x)=1-\cos^2(x)=\left(1-\cos(x)\right)\left(1+\cos(x)\right)\). 3. Cancel the nonzero factor \(1-\cos(x)\). The left side becomes \(1+\cos(x)\), which matches the right side. 4. Therefore, every \(x\) strictly between \(0\) and \(2\pi\) is a solution.

All \(x\) with \(0<x<2\pi\). The excluded values are \(x=0\) and \(x=2\pi\).

- Rewrite \(\sin(2x)\) before deciding whether division is safe. - Moving all terms to one side allows you to factor without discarding a zero factor. - Solve each factor separately and combine the solution sets.

1. Use \(\sin(2x)=2\sin(x)\cos(x)\) and move all terms to one side: \(\sin(x)\left(2\cos(x)-1\right)=0\). 2. The first factor gives \(\sin(x)=0\), so \(x=0\) or \(x=\pi\) on \([0,2\pi)\). 3. The second factor gives \(\cos(x)=\frac{1}{2}\), so \(x=\frac{\pi}{3}\) or \(x=\frac{5\pi}{3}\). 4. Dividing by \(\sin(x)\) would remove the entire \(\sin(x)=0\) branch and lose \(x=0\) and \(x=\pi\).

Dividing by \(\sin(x)\) loses \(x=0\) and \(x=\pi\). The complete solution set is \(x=0,\frac{\pi}{3},\pi,\frac{5\pi}{3}\).

- Replace \(\sin(2x)\) with its double-angle form. - Move all terms to one side so the common cosine factor is visible. - Keep the zero-cosine branch instead of dividing by \(\cos(x)\).

1. Use \(\sin(2x)=2\sin(x)\cos(x)\). Then \(2\sin(x)\cos(x)=\frac{1}{2}\cos(x)\). 2. Move all terms to one side and factor: \(\cos(x)\left(2\sin(x)-\frac{1}{2}\right)=0\). 3. From \(\cos(x)=0\), \(x=\frac{\pi}{2},\frac{3\pi}{2}\). 4. From \(\sin(x)=\frac{1}{4}\), let \(\alpha=\sin^{-1}\left(\frac{1}{4}\right)\). Then \(x=\alpha\) or \(x=\pi-\alpha\). 5. These four values are all in \([0,2\pi)\), and factoring preserves the \(\cos(x)=0\) branch.

\(x=\sin^{-1}\left(\frac{1}{4}\right),\frac{\pi}{2},\pi-\sin^{-1}\left(\frac{1}{4}\right),\frac{3\pi}{2}\)

- Expand the shifted sine and factor the cosine double-angle expression before moving terms. - Preserve the common \(\sin(x)+\cos(x)\) factor rather than dividing by it. - The second factor can be rewritten as one shifted cosine after it is isolated. - Check every branch in the original interval.

1. Expand the left side: \(\sin\left(x+\frac{\pi}{4}\right)=\frac{\sin(x)+\cos(x)}{\sqrt{2}}\). 2. Factor the double-angle expression: \(\cos(2x)=\left(\cos(x)-\sin(x)\right)\left(\cos(x)+\sin(x)\right)\). 3. Multiply by \(\sqrt{2}\) and move all terms to one side: \(\left(\sin(x)+\cos(x)\right)\left(1-\left(\cos(x)-\sin(x)\right)\right)=0\). 4. From \(\sin(x)+\cos(x)=0\), \(\tan(x)=-1\), giving \(x=\frac{3\pi}{4}\) or \(x=\frac{7\pi}{4}\). 5. From \(\cos(x)-\sin(x)=1\), use \(\cos(x)-\sin(x)=\sqrt{2}\cos\left(x+\frac{\pi}{4}\right)\). Then \(\cos\left(x+\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}\). 6. On \([0,2\pi)\), this gives \(x=0\) or \(x=\frac{3\pi}{2}\). 7. Combining the two branches gives all four solutions.

\(x=0,\frac{3\pi}{4},\frac{3\pi}{2},\frac{7\pi}{4}\)

- Represent the linear combination of sine and cosine as one shifted sine function before solving. - Match the two coefficients after expanding the shifted sine form. - After the conversion, solve the basic sine equation for the shifted angle and then translate those solutions back to \(x\). - Check which members of the periodic solution families lie in the original interval.

1. Write \(R\sin(x+\phi)=R\cos(\phi)\sin(x)+R\sin(\phi)\cos(x)\). 2. Matching coefficients gives \(R\cos(\phi)=3\) and \(R\sin(\phi)=4\), so \(R=5\) and \(\phi=\tan^{-1}\left(\frac{4}{3}\right)\approx0.9273\). 3. The equation becomes \(5\sin(x+\phi)=2\), so \(\sin(x+\phi)=0.4\). 4. Let \(\alpha=\sin^{-1}(0.4)\approx0.4115\). Then \(x+\phi=\alpha+2\pi k\) or \(x+\phi=\pi-\alpha+2\pi k\). 5. Restricting to \(0\le x<2\pi\) gives \(x\approx5.7674\) from the first family and \(x\approx1.8028\) from the second. 6. To the nearest hundredth, the solutions are \(x\approx1.80\) and \(x\approx5.77\).

\(3\sin(x)+4\cos(x)=5\sin(x+\phi)\), where \(\phi=\tan^{-1}\left(\frac{4}{3}\right)\). \(x\approx1.80\) or \(x\approx5.77\)