Solve the equation (x in radians and in degrees) for all exact solutions where appropriate. Round approximate answers
in radians to four decimal places and approximate answers in degrees to the nearest tenth.
-$\sin ^ { 2 } x + \sin x = 0$
A) $\left\{ \frac { 3 \pi } { 2 } + n \pi \right\}$
B) $\left\{ \mathrm { n } \pi , \frac { 3 \pi } { 2 } + 2 \mathrm { n } \pi \right\}$
C) $\left\{ \frac { \pi } { 2 } + 2 n \pi \right\}$
D) $\left\{ n \pi , \frac { \pi } { 2 } + 2 n \pi \right\}$

Question

Quizplus · Accepted Answer

To solve the equation $\sin^2x + \sin x = 0$, factor out $\sin x$ to get $\sin x(\sin x + 1) = 0$. This gives two equations: $\sin x = 0$ and $\sin x = -1$. The solutions to $\sin x = 0$ are $x = n\pi$, where $n$ is any integer, because the sine function is zero at integer multiples of $\pi$. The solution to $\sin x = -1$ is $x = \frac{3\pi}{2} + 2n\pi$, because the sine function is -1 at an angle of $\frac{3\pi}{2}$ plus any integer multiple of $2\pi$. Therefore, the correct set of solutions is $\left\{ n\pi, \frac{3\pi}{2} + 2n\pi \right\}$.

Solve the Equation (X in Radians and in Degrees) for All