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) = -\frac12 \text{ or } \sin(3x) = 1`

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

`3x = \sin^(-1)\left(-\frac12\right) + 2n\pi \text{ or } 3x = \pi - \sin^(-1)\left(-\frac12\right) + 2n\pi`

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

`\implies \boxed{x = -\frac\pi{18} + \frac{2n\pi}3} \text{ or } \boxed{x = (7\pi)/(18) + \frac{2n\pi}3}`

and

`3x = \sin^(-1)(1) + 2n\pi`

`\implies 3x = \frac\pi2 + 2n\pi`

`\implies \boxed{x = \frac\pi6 + \frac{2n\pi}3}`

(where
`n` is any integer)