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

Question

Quizplus · Accepted Answer

To solve the equation $\sin^2x + \sin x = 0$, factor it to get $\sin x(\sin x + 1) = 0$. This gives two equations: $\sin x = 0$ and $\sin x = -1$. The solutions in the interval $[0, 2\pi)$ are $x = 0, \pi$ for $\sin x = 0$ and $x = \frac{3\pi}{2}$ for $\sin x = -1$, which matches option A.

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 $[ 0,2 \pi ) .$