3 sin x = 1 + cos 2x

Question

0 is less than or equal to x, which is less than 2 pi. Solve using trigonometric equations. Round to the nearest tenth when necessary.

Answer 1

3 sin x = 1 + cos 2x

use the trig identity: cos(2x) = 1 - 2sin^2(x)

3sin(x) = 1 + 1 - 2sin^2(x)   now treat this as a quadratic equation.

3u = 2 - 2u^2   here u = sin(x)

2u^2 + 3u - 2 = 0

(2u - 1)(u + 2) = 0

u = 1/2, u = -2

sin(x) = 1/2, sin(x) = -2  <--- ignore this last solution, since although it is a solution to the quadratic equation it cannot be a solution to the trig equation, because |sin(x)| <= 1.

sin(x) = 0.5, therefore x = pi/6, and x = 5pi/6