Test Objectives
• Demonstrate the ability to solve a trigonometric equation with multiple angles
• Demonstrate the ability to solve a trigonometric equation with steps inside
Solving Trigonometric Equations Part V Practice Test:

#1:

Instructions: Solve each for 0 ≤ θ < 2π.

$$a)\hspace{.1em}2\text{cos}\hspace{.1em}θ=\text{sin}\hspace{.1em}2θ$$

$$b)\hspace{.1em}2\text{cos}^2 θ - 2=-\text{cos}\hspace{.1em}2θ$$

#2:

Instructions: Solve each for 0 ≤ θ < 2π.

$$a)\hspace{.1em}\text{cos}\hspace{.1em}2θ=-2\text{cos}^2 θ$$

$$b)\hspace{.1em}2\text{sin}\hspace{.1em}2θ=-\sqrt{3}\text{sin}\hspace{.1em}θ + 3\text{sin}\hspace{.1em}2 θ$$

#3:

Instructions: Solve each for 0 ≤ θ < 2π.

$$a)\hspace{.1em}4\text{cos}^2 θ - \text{sin}^2 2θ=0$$

$$b)\hspace{.1em}2\text{cos}^2 θ=1 - \text{cos}\hspace{.1em}2θ$$

#4:

Instructions: Solve each for 0 ≤ θ < 2π.

$$a)\hspace{.1em}-\frac{1}{5}\text{csc}\left(θ + \frac{π}{2}\right)=\frac{2}{5}$$

$$b)\hspace{.1em}-\frac{1}{3}\text{sec}\frac{θ}{2}=\frac{\sqrt{2}}{3}$$

#5:

Instructions: Solve each for 0 ≤ θ < 2π.

$$a)\hspace{.1em}-\text{cos}\left(3θ + \frac{3π}{2}\right)=-\frac{\sqrt{2}}{2}$$

$$b)\hspace{.1em}\sqrt{3}=2\text{cos}\left(\frac{θ}{4}+ \frac{5π}{3}\right)$$

Written Solutions:

#1:

Solutions:

$$a)\hspace{.1em}\left\{\frac{π}{2}, \frac{3π}{2}\right\}$$

$$b)\hspace{.1em}\left\{\frac{π}{6}, \frac{5π}{6}, \frac{7π}{6}, \frac{11π}{6}\right\}$$

#2:

Solutions:

$$a)\hspace{.1em}\left\{\frac{π}{3}, \frac{2π}{3}, \frac{4π}{3}, \frac{5π}{3}\right\}$$

$$b)\hspace{.1em}\left\{0, \frac{π}{6}, π, \frac{11π}{6}\right\}$$

#3:

Solutions:

$$a)\hspace{.1em}\left\{\frac{π}{2}, \frac{3π}{2}\right\}$$

$$b)\hspace{.1em}\left\{\frac{π}{4}, \frac{3π}{4}, \frac{5π}{4}, \frac{7π}{4}\right\}$$

#4:

Solutions:

$$a)\hspace{.1em}\left\{\frac{2π}{3}, \frac{4π}{3}\right\}$$

$$b)\hspace{.1em}\left\{\frac{3π}{2}\right\}$$

#5:

Solutions:

$$a)\hspace{.1em}\left\{\frac{π}{12}, \frac{π}{4}, \frac{3π}{4}, \frac{11π}{12}, \frac{17π}{12}, \frac{19π}{12}\right\}$$

$$b)\hspace{.1em}\left\{\frac{2π}{3}\right\}$$