Question

The number of ordered pair(s) of $(x,y)$satisfying $3y=[\sin x+[\sin x+\left[\sin x\right]\left]\right]$ and $[y+[y\left]\right]=2\cos x$ is
​​​​​​​(correct answer + 1, wrong answer - 0.25)

• A. $2$
• B. $1$
• C. $0$
• D. $3$

Answer 1

The correct option is C $0$

$3y=[\sin x+[\sin x+\left[\sin x\right]\left]\right]\Rightarrow 3y=3\left[\sin x\right]\Rightarrow y=\left[\sin x\right]\cdots \left(1\right)$
Also,
$[y+[y\left]\right]=2\cos x\Rightarrow 2\left[y\right]=2\cos x\Rightarrow \left[y\right]=\cos x$
Using equation $\left(1\right)$, we get
$\left[\sin x\right]=\cos x$

When $\cos x=-1\Rightarrow x=(2n+1)\pi$
But $\sin (2n+1)\pi =0$, so no solution

When $\Rightarrow \cos x=0\Rightarrow x=\frac{(2n+1)\pi }{2}$
But $\sin \left(\frac{(2n+1)\pi }{2}\right)=\pm 1$
So, no solution

When $\cos x=1\Rightarrow x=2n\pi$
But $\sin 2n\pi =0$, so no solution

Hence, there is no pair of $(x,y)$.