Solve the equation in the interval $$\left[ 0 ^ { \circ } , 360 ^ { \circ } \right)$$ ). Give solutions to the nearest tenth, if necessary.
-$\sin 2 \theta = - \sin \theta$

Question

Quizplus · Accepted Answer

The equation is equivalent to $\sin 2 \theta + \sin \theta = 0$, which factors as $\sin \theta (2 \cos \theta + 1) = 0$. So, $\sin \theta = 0$ or $2 \cos \theta + 1 = 0$. If $\sin \theta = 0$, then $\theta = 0^\circ$ or $\theta = 180^\circ$. If $2 \cos \theta + 1 = 0$, then $\cos \theta = -1/2$, which gives $\theta = 120^\circ$ or $\theta = 240^\circ$. Therefore, the solutions are $\left\{ 0^\circ, 120^\circ, 180^\circ, 240^\circ \right\}$.

Solve the Equation in the Interval [0∘,360∘)\left[ 0 ^ { \circ } , 360 ^ { \circ } \right)[0∘,360∘)

Solve the Equation in the Interval $\left[ 0 ^ { \circ } , 360 ^ { \circ } \right)$