Recall the Pythagorean identity,
[tex]\sin^2(x) + \cos^2(x) = 1[/tex]
Use it to rewrite the equation in terms of cos only.
[tex]12 (1 - \cos^2(x)) + \cos(x) - 6 = 0[/tex]
[tex]-12 \cos^2(x) + \cos(x) + 6 = 0[/tex]
Factorize the left side.
[tex]-(3\cos(x) + 2) (4\cos(x) - 3) = 0[/tex]
Solve the two cases for [tex]\cos(x)[/tex].
[tex]3 \cos(x) + 2 = 0 \text{ or } 4\cos(x) - 3 = 0[/tex]
[tex]\cos(x) = -\dfrac23 \text{ or } \cos(x) = \dfrac34[/tex]
From the first equation, we get one family of solutions:
[tex]\cos(x) = -\dfrac23[/tex]
[tex]x = \cos^{-1}\left(-\dfrac23\right) + 2n\pi \text{ or } x = -\cos^{-1}\left(-\dfrac23\right) + 2n\pi[/tex]
From the second equation, we get another family:
[tex]\cos(x) = \dfrac34[/tex]
[tex]x = \cos^{-1}\left(\dfrac34\right) + 2n\pi \text{ or } x = -\cos^{-1}\left(\dfrac34\right) + 2n\pi[/tex]
where [tex]n[/tex] is an integer.
We get solutions in the given domain for
[tex]n = 0 \implies \boxed{x = \cos^{-1}\left(-\dfrac23\right)} \text{ or } \boxed{x = \cos^{-1}\left(\dfrac34\right)}[/tex]
[tex]n=1 \implies \boxed{x = 2\pi - \cos^{-1}\left(-\dfrac23\right)} \text{ or } \boxed{x = 2\pi - \cos^{-1}\left(\dfrac34\right)}[/tex]
