How to prove the identity $3\sin^4x-2\sin^6x=1-3\cos^4x+2\cos^6x$? I'm trying to prove a trigonometric identity but I can't. I've been trying a lot but I can't prove it. The identity says like this:
$$3\sin^4x-2\sin^6x=1-3\cos^4x+2\cos^6x$$
The identity would be easy if $1-\cos^4x=\sin^4x$ and $1-\cos^6x=\sin^6x$ but we know that $\sin^4x+\cos^4x$ isn't equal with $1$ and $\sin^6x+\cos^6x$ isn't equal with $1$. 
Can anybody help me?!
Thank you!

HINT:
As $\sin^2x+\cos^2x=1,$
$\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-2\sin^2x\cos^2x \ \ \ \ (1)$
$\sin^6x+\cos^6x=(\sin^2x+\cos^2x)^3-3\sin^2x\cos^2x(\sin^2x+\cos^2x)$
$\implies \sin^6x+\cos^6x=1-3\sin^2x\cos^2x \ \ \ \ (2)$
Now equate the values of  $\sin^2x\cos^2x$ from $(1),(2)$
