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Find all 4 complex roots to $3z^4-z^3+2z^2-z+3=0$ using a trig substitution

A previous part of the question made us prove that $2\cos n\theta=z^n+z^{-n}$ and it wants us to prove that the polynomial can be expressed as $6\cos 2\theta -2\cos\theta +2=0$. I'm assuming that using $6\cos 2\theta -2\cos\theta +2=0$ we will be able to find all the roots of the equation but I'm unable to see how to get the polynomial into the trig equation and then how to solve the trig equation. Any help would be appreciated.

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Hint: The given equation is equivalent to $$ 3(z^2+\frac{1}{z^2})-(z+\frac{1}{z})+2=0. $$ By letting $z+\frac{1}{z}=w$, we get $$ 3(w^2-2)-w+2=0. $$ Solve this quadratic equation for $w$ and subsequently solve $z+\frac{1}{z}=w$. Since $w$ is real and $|w|\le 2$, we can substitute $z=e^{i\theta}$, $w=2\cos \theta$ and solve for $\theta$.

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Use that $$\cos(2x)=2\cos^2(x)-1$$

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