<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] It is preferable to change the variable so let $t=\cos x$ and we solve the quadratic equation
$$6t^2+t=1$$
and we find $t=\frac13$ and $t=-\frac12$. Now 
$$\cos x=\frac 13\iff \left(x=\arccos\left(\frac13\right)+2k\pi\right)\lor\left( x=2k\pi-\arccos\left(\frac13\right)\right)$$
and 
$$\cos x=-\frac12\iff \left(x=\frac{2\pi}3+2k\pi\right)\lor\left( x=2k\pi-\frac{2\pi}3\right)$$
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