How do i solve Sin(2x)^2-Cos(2x)=Cos(2x) ... so far i've gotten to the step of
4(1-cos^x)(cos^2x)=2Cos(2x) what do i do next?

Question

How do i solve Sin(2x)^2-Cos(2x)=Cos(2x) ... so far i've gotten to the step of    
4(1-cos^x)(cos^2x)=2Cos(2x) what do i do next?

anonymous · Answer

$sin(2x)^2=2\cos 2x$
$sin(2x)^2-2\cos 2x=0$
$1-\cos(2x)^2-2\cos(2x)=0$
$-\cos(2x)^2-2\cos(2x)+1=0$
Now, let $cos(2x)=t$. Then:
$-t^2-2t+1=0$
solve this for $t$
you will get:
$t=\huge \frac{-2\mp2\sqrt{2}}{2}$
now make back to x.
and solve for x for each of t values

anonymous · Answer

@myko  Thank you! i will work it using your method and see what what i get :) Thank you very much!