# Question #da791

Aug 22, 2016

$x = \frac{\pi}{4} + n \frac{\pi}{2} , n \in \mathbb{Z}$

#### Explanation:

$2 {\sin}^{2} \left(x\right) = 1$

$\implies {\sin}^{2} \left(x\right) = \frac{1}{2}$

$\implies \sin \left(x\right) = \pm \sqrt{\frac{1}{2}} = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$

If we look at a unit circle, we will find that $| \sin \left(x\right) | = \frac{\sqrt{2}}{2}$ at an angle of $\frac{\pi}{4}$ in each quadrant.

As we want both positive and negative values for $\sin \left(x\right)$, we find that $\frac{\pi}{4}$ plus any integer multiple of $\frac{\pi}{2}$ is a solution. Thus we get the solution set

$x = \frac{\pi}{4} + n \frac{\pi}{2} , n \in \mathbb{Z}$