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

Solve the following for $\displaystyle x$:

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

$\displaystyle 0 \leq x <2\pi$

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

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- Oct 28th 2010, 03:22 PMqhdusEasy trig question
It's been awhile since I've done high school math, help would be greatly appreciated!

Solve the following for $\displaystyle x$:

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

$\displaystyle 0 \leq x <2\pi$

The solution set is... (should be 4) - Oct 28th 2010, 03:32 PMpickslides
Looks better like this

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

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

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

__Spoiler__: - Oct 28th 2010, 03:49 PMqhdus
Wouldn't it be that

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

I think I got it now though. - Oct 28th 2010, 03:51 PMskeeter
$\displaystyle \displaystyle \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$