Find all solutions in the interval [0, 2π).
-$$\sin ^ { 2 } \left( \frac { x } { 2 } \right) = \sin ^ { 2 } x$$
A) $\frac { 2 \pi } { 3 } , \frac { 4 \pi } { 3 }$
B) $0 , \pi$
C) $0 , \frac { 2 \pi } { 3 }$
D) $0 , \frac { 2 \pi } { 3 } , \frac { 4 \pi } { 3 }$

Question

Quizplus · Accepted Answer

The equation $\sin^2\left(\frac{x}{2}\right) = \sin^2(x)$ is satisfied when $x = 0$, $x = \frac{2\pi}{3}$, and $x = \frac{4\pi}{3}$ within the interval [0, 2π). This is because at these points, the values of $\sin^2\left(\frac{x}{2}\right)$ and $\sin^2(x)$ are equal, either due to the angles being equivalent in their sine values or due to the properties of sine function symmetry and periodicity.

Find All Solutions in the Interval [0, 2π) sin⁡2(x2)=sin⁡2x\sin ^ { 2 } \left( \frac { x } { 2 } \right) = \sin ^ { 2 } xsin2(2x​)=sin2x

Find All Solutions in the Interval [0, 2π) $\sin ^ { 2 } \left( \frac { x } { 2 } \right) = \sin ^ { 2 } x$