Question

The number of solutions of the equations $y=\frac{1}{3}[\sin x+[\sin x+\left[\sin x\right]\left]\right]$ and $[y+[y\left]\right]=2\cos x$, where $[.]$ denotes the greatest integer function is

• A. $0$
• B. $1$
• C. $2$
• D. infinite

Answer 1

The correct option is A $0$

$y=\frac{1}{3}[\sin x+[\sin x+\left[\sin x\right]\left]\right]$
$\Rightarrow y=\frac{1}{3}\left(\right[\sin x]+[\sin x]+[\sin x\left]\right)$
$\Rightarrow y=\left[\sin x\right]$

$[y+[y\left]\right]=2\cos x$
$\Rightarrow 2\left[y\right]=2\cos x$
$\Rightarrow \left[y\right]=\cos x$

Case $\left(1\right)$: When $-1\le \sin x<0$
Then $\left[\sin x\right]=-1$
$\Rightarrow y=-1$
$\Rightarrow [-1]=\cos x$
$\Rightarrow \cos x=-1$
$\Rightarrow \sin x=0$ (impossible)

Case $\left(2\right)$: When $0\le \sin x<1$
Then $\left[\sin x\right]=0$
$\Rightarrow y=0$
$\Rightarrow \cos x=0$
$\Rightarrow \sin x=1$ (impossible)

Case $\left(3\right)$: When $\sin x=1$
Then $y=1$
$\Rightarrow \cos x=1$
$\Rightarrow \sin x=0$ (impossible)

Hence, there is no solution.