Easy trig question

**qhdus** · October 28th 2010, 03:22 PM

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)

**pickslides** · October 28th 2010, 03:32 PM

Looks better like this

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

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

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

Spoiler:

**qhdus** · October 28th 2010, 03:49 PM

Wouldn't it be that

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

I think I got it now though.

**skeeter** · October 28th 2010, 03:51 PM

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

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