Solve the equation for solutions in the interval [0, 2 $$[ 0,2 \pi ) .$$
-$\sin 2 x + \sin x = 0$
A) $\left\{ 0 , \frac { 2 \pi } { 3 } , \pi , \frac { 4 \pi } { 3 } \right\}$
B) $\left\{ \frac { \pi } { 8 } , \frac { 9 \pi } { 8 } \right\}$
C) $\varnothing$
D) $\left\{ \frac { \pi } { 4 } , \frac { 3 \pi } { 4 } , \frac { 5 \pi } { 4 } , \frac { 7 \pi } { 4 } \right\}$

Question

Quizplus · Accepted Answer

Using the identity $\sin 2x = 2\sin x \cos x$, the equation becomes $2\sin x \cos x + \sin x = 0$. Factoring out $\sin x$, we get $\sin x (2\cos x + 1) = 0$. Setting each factor to zero gives the solutions $x = 0, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}$ within the interval $[0, 2\pi)$.

Solve the Equation for Solutions in the Interval [0, 2 [0,2π).[ 0,2 \pi ) .[0,2π).

Solve the Equation for Solutions in the Interval [0, 2 $[ 0,2 \pi ) .$