Help solving sin2x = cos2x between 0 and 2pi

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I'm working on an exam review sheet and I was given the answers to the problems I just don't know how to solve this one. I have tried using identities to solve the problem, but everytime I try I get stuck and seem to be getting nowhere.
 
Try to solve this.
\(\displaystyle \L
\begin{array}{rcl}
\sin (2x) &=& \cos (2x) \\
\sin (2x) &=& \sqrt {1 - \sin ^2 (2x)} \\
\end{array}.\)
 
Hello, RadMad!

Solve \(\displaystyle \,\sin2x\:=\:\cos2x\,\) over \(\displaystyle \,0\,\leq\,x\,\<\,2\pi\)
Click to expand...
Divide both sides by \(\displaystyle \cos2x:\;\;\frac{\sin2x}{\cos2x}\:=\:1\)

Then: \(\displaystyle \,\tan2x\:=\:1\;\;\Rightarrow\;\;2x\:=\:\frac{\pi}{4},\;\frac{5\pi}{4},\;\frac{9\pi}{4},\;\frac{13\pi}{4}\)

Therefore: \(\displaystyle \,x\;=\;\frac{\pi}{8}\,\;\frac{\5\pi}{8},\;\frac{9\pi}{8},\;\frac{13\pi}{8}\)
 
Thank you soroban; I was trying to substitute identities for both sides of the equations 1st.
This helped & gives me a new way to look at these problems.
 
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