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.

It's because there are always two solutions for the equation $\cos x = a$ on $[0,2\pi)$ when $|a|<1$. If $x_1$ is one solution, then $x_2 = 2\pi - x_1$ is another solution. Indeed
$$\cos x_2 = \cos(2\pi - x_1) = \cos(-x_1) = \cos x_1 $$
where the second equality holds since $\cos$ is periodic with period $2\pi$ and the third because it's even function.
EDIT: To address the point Prem made in the comments. In your case you are solving $\cos 2x = a$. If it's expected that $2x\in[0,2\pi)$, then you have two solutions. However, if it's expected that $x\in[0,2\pi)$, and therefore $2x\in [0,4\pi)$, there are four solutions instead. The extra two can be obtained by adding $2\pi$ to the first two, i.e. just using periodicity.
