Solve the equation for exact solutions over the interval [0, 2π).
-$2 \sin ^ { 2 } x = \sin x$

Question

Quizplus · Accepted Answer

To solve $2 \sin ^ { 2 } x = \sin x$, we rearrange the equation to $2 \sin ^ { 2 } x - \sin x = 0$, which can be factored as $\sin x (2 \sin x - 1) = 0$. Setting each factor equal to zero gives us $\sin x = 0$ and $2 \sin x - 1 = 0$. Solving $\sin x = 0$ over the interval [0, 2π) gives $x = 0$ and $x = \pi$. Solving $2 \sin x - 1 = 0$ gives $\sin x = \frac{1}{2}$, which has solutions at $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$ within the given interval. Therefore, the correct set of solutions is $\left\{ 0 , \pi , \frac { \pi } { 6 } , \frac { 5 \pi } { 6 } \right\}$.

Solve the Equation for Exact Solutions Over the Interval [0 2sin⁡2x=sin⁡x2 \sin ^ { 2 } x = \sin x2sin2x=sinx

Solve the Equation for Exact Solutions Over the Interval [0 $2 \sin ^ { 2 } x = \sin x$