Free math worksheets

- Replace the left side with an equivalent expression using the Pythagorean identity. - After obtaining a squared trigonometric equation, consider both signs.

1. Use \(1-\cos^2(x)=\sin^2(x)\). 2. Then \(\sin^2(x)=1\), so \(\sin(x)=1\) or \(\sin(x)=-1\). 3. On \([0,2\pi)\), the solutions are \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\).

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

- 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 \(x=\frac{5\pi}{4}\).

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

- Convert the left side using the Pythagorean identity. - A squared sine value creates two possible sine values. - Use all quadrants that match those two signs.

1. Use the Pythagorean identity to rewrite the equation as \(\sin^2(x)=\frac{1}{4}\). 2. Therefore, \(\sin(x)=\frac{1}{2}\) or \(\sin(x)=-\frac{1}{2}\). 3. On \([0,2\pi)\), the solutions are \(x=\frac{\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6}\).

\(x=\frac{\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6}\)

- Compare the left side with a standard cosine double-angle form. - Track the minus sign when rewriting the expression. - Solve for the doubled angle over its transformed interval.

1. Since \(\cos(2x)=\cos^2(x)-\sin^2(x)\), the left side is \(-\cos(2x)\). 2. The equation becomes \(\cos(2x)=0\). 3. For \(2x\in[0,4\pi)\), the solutions are \(2x=\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},\frac{7\pi}{2}\). 4. Therefore, \(x=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\).

\(x=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\)

- 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 \(x=\frac{2\pi}{3}\).

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

- Rewrite the left side as one cosine function first. - Transform the interval to match the doubled angle. - List every solution in the doubled-angle interval before dividing by \(2\).

1. Use \(2\cos^2(x)-1=\cos(2x)\). 2. The equation becomes \(\cos(2x)=\frac{1}{2}\). 3. For \(2x\in[0,4\pi)\), the solutions are \(2x=\frac{\pi}{3},\frac{5\pi}{3},\frac{7\pi}{3},\frac{11\pi}{3}\). 4. Therefore, \(x=\frac{\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6}\).

\(x=\frac{\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6}\)

- 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 x-intercepts of the graphed expression. - Rewrite \(1-2\sin^2(x)\) as a double-angle expression. - Solve the resulting basic cosine equation over the doubled interval.

1. The graph crosses the x-axis four times, near \(\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\). 2. Use \(1-2\sin^2(x)=\cos(2x)\). 3. The equation becomes \(\cos(2x)=0\). 4. Therefore, \(2x=\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},\frac{7\pi}{2}\), so \(x=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\).

a) Four solutions, near \(\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\) b) \(x=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\)

- 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 \(x=0,\frac{\pi}{3},\pi,\frac{5\pi}{3}\).

- Read the intersection x-values before doing algebra. - Replace \(\sin(2x)\) with its double-angle form. - Move one side to the other and factor rather than dividing by a trigonometric function.

1. The graph shows four intersections on \([0,2\pi)\). 2. Use \(\sin(2x)=2\sin(x)\cos(x)\). Then \(2\sin(x)\cos(x)=\cos(x)\). 3. Factor: \(\cos(x)\left(2\sin(x)-1\right)=0\). 4. From \(\cos(x)=0\), \(x=\frac{\pi}{2},\frac{3\pi}{2}\). 5. From \(\sin(x)=\frac{1}{2}\), \(x=\frac{\pi}{6},\frac{5\pi}{6}\).

a) Four solutions b) \(x=\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6},\frac{3\pi}{2}\)

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

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

- Expand the shifted sine and factor the cosine double-angle expression before moving terms. - Preserve a possible zero factor instead of dividing by it. - One branch is a simple relationship between \(\sin(x)\) and \(\cos(x)\); the other can be rewritten as a shifted cosine. - Check every branch against 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 right side: \(\cos(2x)=\cos^2(x)-\sin^2(x)=\left(\cos(x)-\sin(x)\right)\left(\cos(x)+\sin(x)\right)\). 3. Move to a factored equation: \(\left(\sin(x)+\cos(x)\right)\left(\frac{1}{\sqrt{2}}-\left(\cos(x)-\sin(x)\right)\right)=0\). 4. From \(\sin(x)+\cos(x)=0\), \(\tan(x)=-1\), giving \(x=\frac{3\pi}{4},\frac{7\pi}{4}\). 5. From \(\cos(x)-\sin(x)=\frac{1}{\sqrt{2}}\), use \(\cos\left(x+\frac{\pi}{4}\right)=\frac{\cos(x)-\sin(x)}{\sqrt{2}}\) to get \(\cos\left(x+\frac{\pi}{4}\right)=\frac{1}{2}\). 6. This gives \(x=\frac{\pi}{12}\) or \(x=\frac{17\pi}{12}\) on \([0,2\pi)\). 7. Combining both branches gives all four solutions.

\(x=\frac{\pi}{12},\frac{3\pi}{4},\frac{17\pi}{12},\frac{7\pi}{4}\)