Solving $\sin^2 x+ \sin^2 2x-\sin^2 3x-\sin^2 4x=0$ I am stuck on a trigonometric problem which is as follow:

Solve $\sin^2 x+ \sin^2 2x-\sin^2 3x-\sin^2 4x=0$ for $x$.

these kind of problems tease me always. Please help me. To get a rough idea of how  can I approach such problems.
 A: First linearise with:
$$\sin^2 u=\frac{1-\cos 2u}2.$$
Simplifying yields the equation 
$$\cos 2x+\cos 4x=\cos6x+\cos 8x.$$
Then the factorisation formula:
$$\cos p+\cos q=2\cos\frac{p+q}2\cos\frac{p-q}2 $$
lets you rewrite the equation, after some simplification, as
$$\cos x\cos 3x=\cos x\cos 7x\iff \begin{cases}\cos x=0\quad\text{or}\\\cos 3x=\cos 7x.\end{cases}$$
Solutions to the first equation: $\qquad x\equiv \dfrac\pi2\mod\pi$.
Solutions to the second equation:
$$7x\equiv \pm 3x\mod 2\pi\iff\begin{cases} 4x\equiv 0\mod 2\pi\\10x\equiv 0\mod2\pi\end{cases}\iff\begin{cases} x\equiv 0\mod \dfrac\pi2\\[1ex]x\equiv 0\mod\dfrac\pi5\end{cases} $$
The solutions in the second series are redundant either with the first or the third series. The set of solutions can ultimately be described as:
$$x\equiv \dfrac\pi2\mod\pi,\quad x\equiv 0\mod\dfrac\pi5.
$$
A: HINT:
Using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,
$$\sin^23x-\sin^2x+\sin^24x-\sin^22x=\sin2x(\sin4x+\sin6x)$$
Now use Prosthaphaeresis Formulas  $\sin4x+\sin6x=2\sin5x\cos x$
