<s>[INST] Solve for $x$, $\ x^2+(4x^3-3x )^2=1$, given that $x=\cos\alpha$ I know that it would be:
$\cos^2\alpha+3\cos^2\alpha=1$ by identities
then it would be:
$4\cos^2\alpha=1$
dividing $4$ to both sides:
$\cos^2\alpha=\frac{1}{4}$
square both sides:
$\cos\alpha=\pm\frac{1}{2}$
After this, what should I do?
 [/INST] Plugging $x = \cos \alpha$ we get
$$\cos^2\alpha + \cos^2 3\alpha = 1$$
(since $\cos 3\alpha = 4\cos^3\alpha - 3\cos \alpha$) which implies
$$\sin^2\alpha = \cos^23\alpha \iff (\sin\alpha - \cos3\alpha)(\sin\alpha + \cos3\alpha) = 0$$

Case 1. Let $$\sin\alpha - \cos3\alpha = 0$$
or $$\cos\left(\frac{\pi}{2} - \alpha\right) - \cos3\alpha = 0$$ or $$\sin\left(\frac{\pi}{4} + \alpha\right)\sin\left(\frac{\pi}{4} - 2\alpha\right) = 0.$$
This gives either
$$\sin\left(\frac{\pi}{4} + \alpha\right) = 0$$
or
$$\sin\left(\frac{\pi}{4} - 2\alpha\right) = 0$$
Solce them individually.

Case 2. Let $$\sin\alpha + \cos3\alpha = 0$$
This case is similar to the Case 1.
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