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 $\sin 2x = 2\sin x \cos x$, the given equation can be rewritten as $\sin x \cos 2x + 2\cos x \sin x \cos x = 0$, which simplifies to $\sin x (\cos 2x + 2\cos x \cos x) = 0$. This equation can be solved by setting each factor equal to zero and solving for $x$ in the interval $[0, 2\pi)$, leading to the solutions $x = 0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3}$.

Find the Exact Solutions of the Given Equation in the Interval