1. The problem is to find all solutions to the equation $\sin(4x) = 0$. 2. Recall that $\sin(\theta) = 0$ when $\theta = k\pi$ for any integer $k$. 3. Applying this to our equation, set $4x = k\pi$. 4. Solve for $x$: $$x = \frac{k\pi}{4}$$ where $k$ is any integer. 5. This means there are infinitely many solutions, not just three. 6. For example, if $k=0$, then $x=0$; if $k=1$, then $x=\frac{\pi}{4}$; if $k=2$, then $x=\frac{\pi}{2}$; and so on. 7. So the solutions you mentioned ($x=0$, $x=\frac{\pi}{2}$, and $x=\frac{\pi}{4}$) are indeed solutions, but there are infinitely many more. 8. The key is that the sine function is zero at every integer multiple of $\pi$, and since the argument is $4x$, the solutions are spaced by $\frac{\pi}{4}$. Final answer: $$x = \frac{k\pi}{4}, \quad k \in \mathbb{Z}$$