Question

Solve each equation for 0 ≤ θ 2π -2√2=-4sin 2θ

Answer 1

Answer:

Step-by-step explanation:

Begin by dividing both sides by -4 to get:

$\frac{\sqrt{2} }{2}=sin2\theta$ and then take the inverse sin of both sides to get

$sin^{-1}(\frac{\sqrt{2} }{2})=2\theta$ (the inverse sin undoes the sin on the right). Now we need to look on our unit circle to find the angles where $sin(\frac{\sqrt{2} }{2})$ is positive. There are 2 places. Sin is positive in both QI and QII. The angles are

$\frac{\pi}{4},\frac{3\pi}{4}$. Therefore, our 2 equations are

$\frac{\pi}{4}=2\theta$  and  $\frac{3\pi}{4}=2\theta$.  Solving the first equation:

$\theta=\frac{\pi}{8}$

and the second equation:

$\theta=\frac{3\pi}{8}$