Find all solutions of the following equation in the interval [0, 2$\pi $).
$\sin ^ { 3 } x = \sin x$
A) $x = 0 , \frac { \pi } { 6 } , \frac { 7 \pi } { 6 }$
B) $x = 0 , \frac { \pi } { 4 } , \frac { 7 \pi } { 4 }$
C) $x = 0 , \frac { \pi } { 2 } , \frac { 3 \pi } { 2 }$
D) $x = 0 , \pi$
E) $x = 0 , \frac { \pi } { 2 } , \pi , \frac { 3 \pi } { 2 }$

Question

Find all solutions of the following equation in the interval [0, 2$\pi $). 
 $\sin ^ { 3 } x = \sin x$  
A) $x = 0 , \frac { \pi } { 6 } , \frac { 7 \pi } { 6 }$
B) $x = 0 , \frac { \pi } { 4 } , \frac { 7 \pi } { 4 }$
C) $x = 0 , \frac { \pi } { 2 } , \frac { 3 \pi } { 2 }$
D) $x = 0 , \pi$
E) $x = 0 , \frac { \pi } { 2 } , \pi , \frac { 3 \pi } { 2 }$

Quizplus · Accepted Answer

To solve the equation $\sin^3 x = \sin x$, we can factor it as $\sin x(\sin^2 x - 1) = 0$, which simplifies to $\sin x(\cos^2 x) = 0$ using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$. This gives us two cases to consider: $\sin x = 0$ and $\cos x = 0$. The solutions for $\sin x = 0$ in the interval $[0, 2\pi)$ are $x = 0$ and $x = \pi$. The solutions for $\cos x = 0$ in the interval $[0, 2\pi)$ are $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$. Combining these, we get $x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$.

Find All Solutions of the Following Equation in the Interval π\pi π

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