Finding the number of solutions to $\cos^4(2x)+2\sin^2(2x)=17(1+\sin 2x)^4$ for $x\in(0,2\pi)$ 
Number of solution of the equation
$\cos^4(2x)+2\sin^2(2x)=17(1+\sin 2x)^4\; \forall $ $x\in(0,2\pi)$

what i try
$\cos^4(2x)+2\sin^2 2x=17(1+\sin^2(2x)+2\sin 2x)^2$
$1+\sin^4 (2x)=17(1+\sin^4 2x+2\sin^2 2x+4\sin^24x+4\sin 2x(1+\sin^2 2x))$
$16\sin^4 (2x)+68\sin^3 2x+34\sin^2 2x+68\sin 2x+68\sin^2 4x+16=0$
How do i solve it Help me please

You're not required to find all solutions, just to find how many there are.
Let $u=\sin(2x)$. Then the trigonometric equation in $x$ becomes a polynomial equation in $u$:
$$
0 = (1 - u)^4 + 2 u^2 - 17 (1 + u)^4 = -2 (8 u^4 + 36 u^3 + 47 u^2 + 36 u + 8)
$$
Now plot this function of $u$ and see how many solutions are in the interval $[-1,1]$ so that $u$ can be the sine of something.
The graph below tells us that there is exactly one solution in $[-1,1]$.

