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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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: $\displaystyle x = \frac{\pi}{4}$ and?
Wouldn't it be that $\displaystyle sin(x) = \pm \frac{\sqrt{2}}{2}$? I think I got it now though.
$\displaystyle \displaystyle \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$
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