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**Educated** I'm having a bit of trouble solving this trig problem:

At first when I looked at it it seemed easy but as I tried to solve it, it got really confusing and I got stuck on the $\displaystyle \sin^3(x)$ part.

Here's what I did:

Double angle formula: $\displaystyle \sin(2 \theta) = 2 \sin (\theta) \cos (\theta)$

Double angle formula: $\displaystyle \cos(2 \theta) = 1-2\sin^2(\theta)$

$\displaystyle 0=\sin(4x) - \cos(x)$

$\displaystyle 0=2\cdot\sin(2x)\cdot\cos(2x) - \cos(x)$

$\displaystyle 0=4[sin(x)\cos(x)]\cdot[1-\sin^2(x)] - \cos(x)$

$\displaystyle 0=4\sin(x)\cos(x) - 8\sin^3(x)\cos(x) - \cos(x)$

$\displaystyle 0=\cos(x)[4\sin(x) - 8\sin^3(x) - 1]$

Now I can get the general solution for the $\displaystyle \cos(x)$ part but what about the $\displaystyle 4\sin(x) - 8\sin^3(x) - 1$ part? How would I solve that?

EDIT: Sorry I just realised I placed this in the wrong section :/

Can a mod please move it to the trig section?