Question

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

Answer 1

$\sin x\cos y=1$

$\Rightarrow \cos y=\frac{1}{\sin x}$

$\Rightarrow \cos y=\csc x.....\left(1\right)$

There are only two values which satisfies equation $\left(1\right)$, i.e., $1$ and $-1$.

Case I:- For $1$

$\cos y=\csc x=1$

$\because x,y\in [0,2\pi ]$

Therefore,

$y=0,2\pi$

$x=\frac{\pi }{2}$

Hence the ordered pair will be $(\frac{\pi }{2},0)$ and $(\frac{\pi }{2},2\pi )$.

Case II:- For $-1$

$\cos y=\csc x=-1$

$\because x,y\in [0,2\pi ]$

Therefore,

$x=\pi$

$y=\frac{3\pi }{2}$

Hence the ordered pair will be $(\pi ,\frac{3\pi }{2})$.

Hence there are three ordered pairs for which $x,y\in [0,2\pi ]$ and these pairs are $(\frac{\pi }{2},0),(\frac{\pi }{2},2\pi )$ and $(\pi ,\frac{3\pi }{2})$.