algebra precalculus - How to prove the identity $3sin^4x-2sin^6x=1-3cos^4x+2cos^6x$?

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!

Answer

$$3\sin^4 x - 2\sin^6 x = 3 (\sin^2 x)^2 - 2(\sin^2x)^3 = (\sin^2 x)^2(3 - 2\sin^2 x)\cdots$$

Now we can use the Pythagorean Identity: $$\sin^2x + \cos^2x = 1 \iff \sin^2 x = 1 -\cos^2 x$$

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$$\begin{align} 3\sin^4 x - 2\sin^6 x & = 3 (\sin^2 x)^2 - 2(\sin^2 x)^3 \\ \\
& = (\sin^2x)^2 (3 - 2\sin^2 x) \\ \\
& = (1 - \cos^2 x)^2\Big(3 - 2(1 - \cos ^2 x)\Big)\\ \\
& = (1- 2\cos^2 x + \cos^4 x)(1 + 2 \cos^2 x) \\ \\
& = 1 - 3 \cos^4 x + 2 \cos^6 x\end{align}$$