Question

If $x,y\in [0,2\pi ]$, then find the total number of ordered pairs $(x,y)$ satisfying the equation $\sin x\cos y=1$

Answer 1

$\sin x\cos y=1$

Possible ways are $\Rightarrow$ $\sin x=1$, $\cos y=1$ or $\sin x=-1$, $\cos y=-1$

If $\sin x=1$, $\cos y=1$; hence, $x=\pi /2$, $y=0,2\pi$

If $\sin x=-1$, $\cos y=-1$; hence, $x=3\pi /2$, $y=\pi$

Thus, the possible ordered pairs are $(\frac{\pi }{2},0)$, $(\frac{\pi }{2},2\pi )$ and $(\frac{3\pi }{2},\pi )$
$\therefore$ Total number of ordered pairs is $\Rightarrow 3$