1. ## Easy trig question

It's been awhile since I've done high school math, help would be greatly appreciated!

Solve the following for $x$:

$2sin^2(x)=1$

$0 \leq x <2\pi$

The solution set is... (should be 4)

2. Looks better like this

$2\sin^2(x)=1$

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

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

Spoiler:
$x = \frac{\pi}{4}$ and?

3. Wouldn't it be that

$sin(x) = \pm \frac{\sqrt{2}}{2}$?

I think I got it now though.

4. $\displaystyle \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$