## Precalculus (6th Edition) Blitzer

$$x=\frac{\pi }{3}, \qquad x=\frac{2\pi }{3}, \\ x= \frac{4\pi }{3}, \qquad x=\frac{5\pi }{3}$$
We solve as follows: $$4\sin ^2 x -3=0 \quad \Rightarrow \quad \sin ^2 x = \frac{3}{4}\quad \Rightarrow \quad \sin x = \pm \frac{\sqrt{3}}{2} \quad \Rightarrow \quad x=\frac{\pi }{3}+ 2n\pi, \quad \text{ or } \quad x=\frac{2\pi }{3} + 2n \pi, \quad \text{ or } \quad x=\frac{4\pi }{3}+ 2n \pi \quad \text { or } \quad x=\frac{5\pi }{3} \quad n \in \mathbb{Z}$$So, $x= \frac{\pi }{3}$, $x= \frac{2\pi }{3}$, $x= \frac{4\pi }{3}$, and $x=\frac{5\pi }{3}$ are the only solutions in the interval $[0, 2\pi )$.