<s>[INST] Evaluate $\sin^2x = \frac{2+ \sqrt{3}}{4}$ 
Evaluate $\sin^2x = \frac{2+ \sqrt{3}}{4}$

Find value of $2x$
I worked it out as such:
$\sin^2x = \frac{2+ \sqrt{3}}{4} \implies \frac{1- \cos 2x}{2} = \frac{2+ \sqrt{3}}{4}$
$2 - 2 \cos 2x = 2 + \sqrt{3}$
$- 2 \cos 2x = + \sqrt{3}$
$ \cos 2x = \frac{- \sqrt{3}}{2}$
And $ 2x = \frac{5\pi}{6}$
I was told that the angle $2x$ has another value: $2x = 2\pi - \frac{5\pi}{6} = \frac{7\pi}{6}$
Why is there another value and what is this value for or mean in this question context.
 [/INST] The reason why you have two solutions is because $\sin^2(x) = \sin^2(2\pi - x)$
therefore both solutions work.
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