Solve the equation on the interval $0 \leq \theta

Question

Solve the equation on the interval $0 \leq \theta < 2 \pi .$
-$\sin ^ { 2 } \theta + \sin \theta = 0$
A) $0 , \pi , \frac { \pi } { 3 } , \frac { 2 \pi } { 3 }$
B) $0 , \pi , \frac { 3 \pi } { 2 }$
C) $0 , \pi , \frac { 4 \pi } { 3 } , \frac { 5 \pi } { 3 }$
D) $0 , \pi , \frac { \pi } { 3 } , \frac { 5 \pi } { 3 }$

Quizplus · Accepted Answer

This equation can be factored as $\sin \theta (\sin \theta + 1) = 0$, leading to $\sin \theta = 0$ or $\sin \theta = -1$. The solutions within the given interval are $\theta = 0, \pi$ for $\sin \theta = 0$ and $\theta = \frac{3\pi}{2}$ for $\sin \theta = -1$.

Solve the Equation on the Interval 0≤θ<2π.0 \leq \theta < 2 \pi .0≤θ<2π.

Solve the Equation on the Interval $0 \leq \theta < 2 \pi .$