<s>[INST] Prove this identity: $\frac{2\sin^4x+\cos^2x-2\cos^4x}{3\sin^2x-1} =1$ I am stuck on this identity
$$\frac{2\sin^4x+\cos^2x-2\cos^4x}{3\sin^2x-1} =1$$
I began working on the left side trying to get things to cancel out or equal one by the Pythagorean identities. I am stuck and can't get it to reduce anymore. 
 [/INST] Let's take this one step at a time.
First, move the denominator to the right side.
1)  $$ 2sin^4(x) + cos^2(x) - 2cos^4(x) = 3sin^2(x) - 1 $$
Use the identity $ sin^2(x) = 1-cos^2(x) $ to rewrite $ sin^4(x) = (1-cos^2(x))^2$.
2)  $$ 2(1-cos^2(x))^2 + cos^2(x) - 2cos^4(x) = 3sin^2(x) - 1 $$
Now foil out $2(1-cos^2(x))^2 = 2(1 - 2cos^2(x) + cos^4(x))$ and distribute the 2.
3)  $$ 2-4cos^2(x) + 2cos^4(x) + cos^2(x) - 2cos^4(x) = 3sin^2(x) - 1$$
Now there are three things we can do here
First we have $ 2cos^4(x) $ and $ -2cos^4(x) $ so those will cancel out.  Second we have $ -4cos^2(x) + cos^2(x) = -3cos^2(x)$.  And last we can move the 2 from the left side of the equation to the right. 
4)  $$ -3cos^2(x) = 3sin^2(x) - 3 $$
Now factor out the 3 and it cancels out
5)  $$-cos^2(x) = sin^2(x) - 1 $$
And just rearrange it and we get
6)  $$ 1 = sin^2(x) + cos^2(x) $$
Hope that helps!
</s>