Chapter 6 - Section 6.3 - Trigonometric Equations Involving Multiple Angles - 6.3 Problem Set - Page 340: 35

$$x= \frac{ \pi}{8}+ \frac{ k\pi}{2} \ \text{or}\ x= \frac{3 \pi}{8}+ \frac{ k\pi}{2}$$

Given $$ \sin^2 4x = 1 $$ Then $$ \sin 4x =\pm 1 $$ We have two cases for $\sin 4x$ Case 1 $$\sin 4x=1$$ Then $$ 4x=\frac{ \pi}{2}+2k\pi \ \Rightarrow x= \frac{ \pi}{8}+ \frac{ k\pi}{2}$$ Case 2 $$\sin 4x=-1$$ Then $$ 4x=\frac{3 \pi}{2}+2k\pi \ \Rightarrow x= \frac{3 \pi}{8}+ \frac{ k\pi}{2}$$