Find the exact solutions of the given equation in the interval $[ 0,2 \pi )$.
$\sin x \cos 2 x + \cos x \sin 2 x = 0$
A) $x = \frac { \pi } { 2 } , \frac { 7 \pi } { 6 } , \frac { 11 \pi } { 6 }$
B) $x = 0 , \frac { \pi } { 3 } , \frac { 2 \pi } { 3 } , \pi , \frac { 4 \pi } { 3 } , \frac { 5 \pi } { 3 }$
C) $x = 0 , \frac { \pi } { 3 } , \frac { 2 \pi } { 3 } , \pi , \frac { 4 \pi } { 3 }$
D) $x = 0 , \frac { \pi } { 3 } , \pi , \frac { 5 \pi } { 3 }$
E) $x = 0$

Question

Quizplus · Accepted Answer

Using the trigonometric identity $\sin(a)\cos(b) + \cos(a)\sin(b) = \sin(a + b)$, the given equation becomes $\sin(x + 2x) = \sin(3x) = 0$. The solutions to $\sin(3x) = 0$ in the interval $[0, 2\pi)$ are $x = 0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3}$.

Find the Exact Solutions of the Given Equation in the Interval