Question

Sin 3x + 2cos^2 (3x) = 1

Answer 1

Recall the Pythagorean identity,

$\sin^2(x) + \cos^2(x) = 1$

Then we can rewrite the equation as

$\sin(3x) + 2 \cos^2(3x) = 1$

$\sin(3x) + 2 (1 - \sin^2(3x)) = 1$

$1 + \sin(3x) - 2 \sin^2(3x) = 0$

Factorize the left side.

$(1 + 2 \sin(3x)) (1 - \sin(3x)) = 0$

Then we have

$1 + 2\sin(3x) = 0 \text{ or } 1 - \sin(3x) = 0$

$\sin(3x) = -\dfrac12 \text{ or } \sin(3x) = 1$

Solve for $x$. We get two families of solutions:

$3x = \sin^{-1}\left(-\dfrac12\right) + 2n\pi \text{ or } 3x = \pi - \sin^{-1}\left(-\dfrac12\right) + 2n\pi$

$\implies 3x = -\dfrac\pi6 + 2n\pi \text{ or } 3x = \dfrac{7\pi}6 + 2n\pi$

$\implies \boxed{x = -\dfrac\pi{18} + \dfrac{2n\pi}3} \text{ or } \boxed{x = \dfrac{7\pi}{18} + \dfrac{2n\pi}3}$

and

$3x = \sin^{-1}(1) + 2n\pi$

$\implies 3x = \dfrac\pi2 + 2n\pi$

$\implies \boxed{x = \dfrac\pi6 + \dfrac{2n\pi}3}$

(where $n$ is any integer)