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

Wrong expansion jacky
$$16\sin^42x+68\sin^32x+102\sin^22x+68\sin 2x+16=0$$
$$16(\sin 2x+\csc 2x)^2+68(\sin 2x+\csc 2x)+70=0$$
$$\sin 2x +\csc 2x=-15.0479,-1.9521$$
Rejecting $-1.9521$ (Why)
Thus, $$\sin^22x+(15.0479)\sin 2x+1=0$$
Both the roots of the quadratic are negative and since their product is $1$ only one root will lie on the interval $(-1,0)$
So there should be $4$ solutions
