Solve for $x,y$ $\begin{cases}\sin^2x+\sin^2y=\frac{3}{4}\\x+y=\frac{5\pi}{12}\end{cases}$ 
Designate $x,y\in\left(0,\frac{\pi}{2}\right)$ which fulfills
$\begin{cases}\sin^2x+\sin^2y=\frac{3}{4}\\x+y=\frac{5\pi}{12}\end{cases}$

My proof:
$\cos2x=1-2\sin^2x\Rightarrow \sin^2x=\frac{1}{2}-\frac{1}{2}\cos2x$
$\sin^2x+\sin^2y=\frac{3}{4}\Rightarrow1-\frac{1}{2}\left(\cos2x+\cos2y\right)=\frac{3}{4}\Rightarrow \cos2x+\cos2y=\frac{1}{2}\\ \cos x+\cos y=2\cos\frac{x+y}{2}\cos\frac{x-y}{2}\\\cos2x+\cos2y=\frac{1}{2}\Rightarrow \cos(x+y)\cos(x-y)=\frac{1}{4}\\\cos\left(\frac{5\pi}{12}\right)\cos\left(2x-\frac{5\pi}{12}\right)=\frac{1}{4}\\\cos2x=2\cos^2x-1\Rightarrow\cos\left(\frac{10\pi}{12}\right)=2\cos^2\left(\frac{5\pi}{12}\right)-1\Rightarrow\cos\left(\frac{5\pi}{12}\right)=\sqrt{\frac{-\frac{\sqrt{3}}{2}+1}{2}}=\frac{\sqrt{2-\sqrt{3}}}{2}\\\cos\left(2x-\frac{5\pi}{12}\right)=\frac{1}{4}\cdot\frac{2}{\sqrt{2-\sqrt{3}}}=\frac{1}{2\sqrt{2-\sqrt{3}}}$
I don't know what to do next,I don't know if it makes sense what I wrote
 A: From $\sin^2x+\sin^2y=\frac{3}{4}$, we get
$$\cos 2x +\cos 2y = \frac 12$$
or
$$\cos(x+y)\cos(x-y) = \frac14 \implies 
\cos \frac {5\pi}{12}\cos(x-y) =\frac12\sin\frac\pi{6}$$
which leads to,
$$\cos(x-y) = \frac{\sin\frac\pi6}{2\cos\frac{5\pi}{12}}=\frac{2\sin\frac\pi{12}\cos\frac\pi{12}}{2\sin\frac\pi{12}}=\cos\frac\pi{12}$$
Thus,
$$x-y = \frac {\pi}{12}, \>\>\>\>\>x+y=\frac{5\pi}{12}$$
and
$$x= \frac\pi4,\>\>\>\>\>y= \frac\pi6$$
A: I think this is a question about mental arihmetic, which $x=\pi/6, y= \pi/4$.
Or proceed from third step, $\cos2x+\cos2y=2\cos(x+y)(\cos(x-y))=1/2$
hence $2\cos(5\pi/12)(\cos(x-y))=1/2$
then using calculator, or as follows, you compute $\cos(x-y)=\dfrac{\sqrt{2}(\sqrt{3}+1)}{4}$ (note that $\cos(5\pi/12)$=...
Either calculator, or by remember that $\cos(x-y)=\cos(\pi/12)$, 
then solving equation gives the answer
$\\ \\\\\\\\\\ \\\ \\\ \\\ $
Or by direct computation, gives $1-\cos(2y)^2 \dfrac{(\sqrt{3}-1)^4}{4}=\cos(2y)^2+2\cos(2y)+1$, hence $\cos(2y)=0$, giving answer.
