<s>[INST] Solving a quadratic trigonometric equation? The equation is $6 \cos^2x+\cos x=1$,
My work:
$6x^2+x-1=0$
$(3x-1)(2x+1)$
$3x-1=0 ∨ 2x+1=0$
$x=\frac{1}{3} ∨ x= \frac{-1}{2}$
But I do not know how to progress further.
 [/INST] I'd set a different variable to $\cos(x)$ other than $x$ so you don't get them mixed up (like $\alpha$ or something).
You can see from your working that
$$\cos(x)=\frac{1}{3} \Rightarrow x= \cos^{-1}\left(\frac{1}{3}\right)$$
$$\therefore \quad x  \approx 1.23+2\pi n, \quad x \approx5.05+2\pi n$$
$$\cos(x)=-\frac{1}{2} \Rightarrow x= \cos^{-1}\left(-\frac{1}{2}\right)$$
$$\therefore \quad x =\frac{2\pi}{3}+2\pi n, \quad x=\frac{4\pi}{3}+2\pi n$$
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