Solve the equation for exact solutions over the interval [0, 2 $$[ 0,2 \pi )$$
-$2 \sin ^ { 2 } x = \sin x$
A) $\left\{ 0 , \pi , \frac { \pi } { 6 } , \frac { 5 \pi } { 6 } \right\}$
B) $\left\{ \frac { \pi } { 2 } , \frac { 3 \pi } { 2 } , \frac { \pi } { 3 } , \frac { 2 \pi } { 3 } \right\}$
C) $\left\{ \frac { \pi } { 3 } , \frac { 2 \pi } { 3 } \right\}$
D) $\left\{ \frac { \pi } { 6 } , \frac { 5 \pi } { 6 } \right\}$

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 $\sin x = 0$ or $\sin x = \frac{1}{2}$. Solving these gives $x = 0, \pi$ for $\sin x = 0$ and $x = \frac{\pi}{6}, \frac{5\pi}{6}$ for $\sin x = \frac{1}{2}$, all within the interval $[0, 2\pi)$.

Solve the Equation for Exact Solutions Over the Interval [0 [0,2π)[ 0,2 \pi )[0,2π)

Solve the Equation for Exact Solutions Over the Interval [0 $[ 0,2 \pi )$