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?

As mentioned by op, using $\ (a-b)^2=a^2-2ab+b^2$ to solve for the values of x, given the equation $\ cos^2\alpha+\cos^23\alpha=1$ is:
$\pm\frac{\sqrt 2}{2}, \pm0.9238..., \pm0.3826...$
