<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 would not use the same letter, $x$, to refer to two different things.
$$
6\cos^2 x + \cos x - 1 = 0
$$
$$
6u^2 + u - 1 = 0
$$
$$
(3u-1)(2u+1)=0
$$
$$
u = \frac 1 3\text{ or } u = \frac{-1}{2}
$$
$$
\cos x=\frac 1 3 \text{ or }\cos x = \frac{-1}2
$$
$$
\Big(x = \left(\arccos\frac 1 3\right) + 2\pi n \text{ or } \left(\pi-\arccos\frac 1 3\right) + 2\pi n \Big)\text{ or }\Big( x = \frac{2\pi}3+2\pi n \text{ or }x=\frac{4\pi}3+2\pi n \Big)
$$
</s>