Solving $2\cot 2x\cos2x = 1-\sin 2x$ How would I solve the following trigonometric equation? 
$$2\cot 2x\cos2x = 1-\sin 2x$$
I got to this stage: $2\cos^22x = \sin2x - \sin^22x$
How do I continue?
 A: $$2cot2xcos2x=1-sin2x\\ \frac { \cos ^{ 2 }{ 2x }  }{ \sin { 2x }  } =\frac { 1-\sin { 2x }  }{ 2 } \\ 2\left( 1-\sin ^{ 2 }{ 2x }  \right) +\sin ^{ 2 }{ 2x } -\sin { 2x } =0\\ \sin ^{ 2 }{ 2x } +\sin { 2x } -2=0\\ \sin { 2x } =\frac { -1\pm 3 }{ 2 } \\ \sin { 2x\neq -2 } ,\sin { 2x } =1\\ \sin { 2x } =1\Rightarrow 2x=\frac { \pi  }{ 2 } +2n\pi \Rightarrow \\$$
so the final answer is:

$$ x=\frac { \pi  }{ 4 } +n\pi $$

A: How about this:
$$  \frac{2\cos 2x}{\sin 2x}\cos 2x = 1 -\sin 2x$$
$$ 2\cos^2 2x = \sin 2x - \sin^2 2x$$
Use the fact that $\cos^2 2x = 1 -\sin^2 2x$  to express the equation in terms of $\sin 2x$
$$ \sin^2 2x + \sin 2x -2 = 0$$
$$ (\sin 2x +2)(\sin 2x - 1) = 0$$
I'm sure you can finish it from here.
A: That leads to: $2(1-t^2) = t^3-t^4, t = \sin(2x)$. Can you see common factor here?
A: after applying the addition formulas we get
$$2\cos(x)^4-2\cos(x)^2-\sin(x)\cos(x)+1=0$$
we converting this in $\tan(x/2)$ we get$$\left( {t}^{4}-2\,{t}^{3}+2\,{t}^{2}+2\,t+1 \right)  \left( {t}^{2}+2
\,t-1 \right) ^{2}
=0$$ with $$t=\tan(x/2)$$
A: Let $u=\cos 2x$. The equation becomes:
$$2\frac{u^2}{\sqrt{1-u^2}}=1-\sqrt{1-u^2}$$
or
$$2u^2=\sqrt{1-u^2}-1+u^2$$
which leads to
$$(u^2+1)^2=1-u^2$$
that is
$$u^4+3u^2=0$$
which has only one solution: $u=0$, that is, $x\equiv \pi/4\pmod{\pi/2}$
