<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 should be: $$\eqalign{6\cos^2x+\cos x-1=0
&\iff
(3\cos x-1)(2\cos x+1)=0
\\
&\iff 3\cos x-1=0,\;\; 2\cos x+1=0
\\
&\iff\cos x=\tfrac13,\;\; \cos x=-\tfrac12.}$$
Can you take it from here?
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