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?

Answer 1

\(sin(2x)^2=2\cos 2x\)
\(sin(2x)^2-2\cos 2x=0\)
\(1-\cos(2x)^2-2\cos(2x)=0\)
\(-\cos(2x)^2-2\cos(2x)+1=0\)
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

Answer 2

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