<s>[INST] 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?
 [/INST] $$2cot2xcos2x=1-sin2x\\ \frac { \cos ^{ 2 }{ 2x }  }{ \sin { 2x }  } =\frac { 1-\sin { 2x }  }{ 2 } \\ 2\left( 1-\sin ^{ 2 }{ 2x }  \right) +\sin ^{ 2 }{ 2x } -\sin { 2x } =0\\ \sin ^{ 2 }{ 2x } +\sin { 2x } -2=0\\ \sin { 2x } =\frac { -1\pm 3 }{ 2 } \\ \sin { 2x\neq -2 } ,\sin { 2x } =1\\ \sin { 2x } =1\Rightarrow 2x=\frac { \pi  }{ 2 } +2n\pi \Rightarrow \\$$
so the final answer is:

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

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