1. The problem is to solve the equation $\sin^2(x) = 0$. 2. Recall that $\sin^2(x) = (\sin(x))^2$, so the equation means $\sin(x) = 0$. 3. The sine function equals zero at integer multiples of $\pi$, so the general solution is: $$x = n\pi \quad \text{where } n \in \mathbb{Z}$$ 4. This means $x$ can be $0, \pi, 2\pi, -\pi, -2\pi$, and so on. 5. Therefore, all solutions to $\sin^2(x) = 0$ are given by $x = n\pi$ for any integer $n$.