Solve the equation on the interval [0, 2π).
-$2 \sin ^ { 2 } x = \sin x$
A) $0 , \pi , \frac { \pi } { 6 } , \frac { 5 \pi } { 6 }$
B) $\frac { \pi } { 2 } , \frac { 3 \pi } { 2 } , \frac { \pi } { 3 } , \frac { 2 \pi } { 3 }$
C) $\frac { \pi } { 6 } , \frac { 5 \pi } { 6 }$
D) $\frac { \pi } { 3 } , \frac { 2 \pi } { 3 }$

Question

Quizplus · Accepted Answer

First, rewrite the equation as $2\sin^2x - \sin x = 0$. Factoring gives $\sin x(2\sin x - 1) = 0$, leading to $\sin x = 0$ and $\sin x = \frac{1}{2}$. Solving these on the interval [0, 2π) gives $x = 0, \pi$ for $\sin x = 0$ and $x = \frac{\pi}{6}, \frac{5\pi}{6}$ for $\sin x = \frac{1}{2}$.

Solve the Equation on the Interval [0, 2π) 2sin⁡2x=sin⁡x2 \sin ^ { 2 } x = \sin x2sin2x=sinx

Solve the Equation on the Interval [0, 2π) $2 \sin ^ { 2 } x = \sin x$