Suppose $X$ has a uniform distribution on the interval $(0,2\pi)$. Consider $Y=\sin^2(X)$. Then
$$P(Y \le y) = P(X \le x_1) + P(x_2 \le X \le x_3) + P(X \ge x_4).$$
From the symmetry of the function $\sin^2(X)$ and the fact that $X$ has uniform distribution, we have $P(X \le x_1)=P(X \ge x_4)$ and $P(x_2 \le X \le x_3)=2P(x_2 \le X \le \pi)$, so $P(Y \le y) =2P(X \le x_1)+2P(x_2 \le X \le \pi)$, where $x_1$ and $x_2$ are the two solutions to $\sin^2(x)=y,0<x<\pi$.
(Casella, G., & Berger, R. L. Statistical inference. Example 2.1.2)
I feel doubt that the above steps used "the fact that $X$ has uniform distribution". I think the above only used the function $\sin^2(X)$ is symmetric around $\pi$. Am I right?