Finding $Pr(Y_1 > c, Y_2 > c) = 0$ when $Y_1$ and $Y_2$ are sin functions

Question

Let $Y_1 = \sin(2\pi Z)$ and $Y_2 = \sin(4\pi Z)$, where $Z$ is a random variable uniformly distributed between 0 and 1.

I have been asked to show that $Pr(Y_1 > c, Y_2 > c) = 0$ where $c = \sin(\pi/3) = \sqrt{3}/2$.

\begin{align} \Pr(Y_1 > c, Y_2 > c) & = \Pr(\sin(2\pi Z) > c, \sin(4\pi Z) > c)\\ & = \Pr(\frac{1}{6} < Z < \frac{1}{3}, \frac{1}{12} < Z < \frac{5}{12})\\ & = \Pr(\frac{1}{6} < Z < \frac{1}{3})\\ & = \frac{1}{6} \end{align}

Am I doing something wrong in here?

Answer 1

$$\sin (4 \pi Z) > c \iff Z \in (\frac{1}{12}, \frac{1}{6}) \cup (\frac{7}{12}, \frac{2}{3})$$