Question

For the interval $[-2\pi ,2\pi ],\sin x>0$ in the interval:

• A. $(-2\pi ,-\pi )\cup (0,\pi )$
• B. $[-\frac{\pi }{2},\frac{\pi }{2}]$
• C. $(-2\pi ,2\pi )-\left\{0\right\}$
• D. $[-2\pi ,-\frac{\pi }{2}]\cup [\frac{\pi }{2},2\pi ]$

Answer 1

The correct option is A $(-2\pi ,-\pi )\cup (0,\pi )$

From the graph of $\sin x$,

We see that $\sin x>0\forall (-2\pi ,-\pi )\cup (0,\pi )$