Recall the Pythagorean identity,
[tex]\sin^2(x) + \cos^2(x) = 1[/tex]
Then we can rewrite the equation as
[tex]\sin(3x) + 2 \cos^2(3x) = 1[/tex]
[tex]\sin(3x) + 2 (1 - \sin^2(3x)) = 1[/tex]
[tex]1 + \sin(3x) - 2 \sin^2(3x) = 0[/tex]
Factorize the left side.
[tex](1 + 2 \sin(3x)) (1 - \sin(3x)) = 0[/tex]
Then we have
[tex]1 + 2\sin(3x) = 0 \text{ or } 1 - \sin(3x) = 0[/tex]
[tex]\sin(3x) = -\dfrac12 \text{ or } \sin(3x) = 1[/tex]
Solve for [tex]x[/tex]. We get two families of solutions:
[tex]3x = \sin^{-1}\left(-\dfrac12\right) + 2n\pi \text{ or } 3x = \pi - \sin^{-1}\left(-\dfrac12\right) + 2n\pi[/tex]
[tex]\implies 3x = -\dfrac\pi6 + 2n\pi \text{ or } 3x = \dfrac{7\pi}6 + 2n\pi[/tex]
[tex]\implies \boxed{x = -\dfrac\pi{18} + \dfrac{2n\pi}3} \text{ or } \boxed{x = \dfrac{7\pi}{18} + \dfrac{2n\pi}3}[/tex]
and
[tex]3x = \sin^{-1}(1) + 2n\pi[/tex]
[tex]\implies 3x = \dfrac\pi2 + 2n\pi[/tex]
[tex]\implies \boxed{x = \dfrac\pi6 + \dfrac{2n\pi}3}[/tex]
(where [tex]n[/tex] is any integer)
