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

Question

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

Quizplus · Accepted Answer

First, rewrite the equation as $2\sin^2\theta - \sin\theta = 0$. Factoring gives $\sin\theta(2\sin\theta - 1) = 0$, leading to $\sin\theta = 0$ and $\sin\theta = \frac{1}{2}$. Solving these gives $\theta = 0, \pi$ for the first equation and $\theta = \frac{\pi}{6}, \frac{5\pi}{6}$ for the second, covering all solutions in D.

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

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