# How do you solve 2sin^2x=1?

Jun 10, 2015

$S = \left\{x | x = \frac{\pi}{4} + \frac{k}{2} \pi \text{ where } k \in \mathbb{R}\right\}$.

#### Explanation:

$2 {\sin}^{2} x = 1$
${\sin}^{2} x = \frac{1}{2}$
$\sin x = \pm \sqrt{\frac{1}{2}}$
$\sin x = \pm \frac{\sqrt{2}}{2}$
This is a common value that we get with $x = \frac{\pi}{4} , \frac{3}{4} \pi , \frac{5}{4} \pi , \frac{7}{4} \pi , \ldots$
So, the solutions is:
$S = \left\{x | x = \frac{\pi}{4} + \frac{k}{2} \pi \text{ where } k \in \mathbb{R}\right\}$.