Answer:
Step-by-step explanation:
Begin by dividing both sides by -4 to get:
[tex]\frac{\sqrt{2} }{2}=sin2\theta[/tex] and then take the inverse sin of both sides to get
[tex]sin^{-1}(\frac{\sqrt{2} }{2})=2\theta[/tex] (the inverse sin undoes the sin on the right). Now we need to look on our unit circle to find the angles where [tex]sin(\frac{\sqrt{2} }{2})[/tex] is positive. There are 2 places. Sin is positive in both QI and QII. The angles are
[tex]\frac{\pi}{4},\frac{3\pi}{4}[/tex]. Therefore, our 2 equations are
[tex]\frac{\pi}{4}=2\theta[/tex] and [tex]\frac{3\pi}{4}=2\theta[/tex]. Solving the first equation:
[tex]\theta=\frac{\pi}{8}[/tex]
and the second equation:
[tex]\theta=\frac{3\pi}{8}[/tex]
