Solve $y=\frac{1}{3}[\sin x+[\sin x+[\sin x]]]$ & $[y+[y]]=2\cos x$

Question

$[x]$ represents the greatest integer function $y=\frac{1}{3}[\sin x+[\sin x+[\sin x]]]$

&

$[y+[y]]=2 \cos x$

Find the number of solution

My approach is as follows

$\sin x \in (\pi,2\pi)$

$y=\frac{1}{3}[\sin x+[\sin x+[\sin x]]]$

$y=-1$

$[-1+[-1]]=2 \cos x$

$\cos x=-1$ which is possible at $x=\pi$ hence NO SOLUTION

$\sin x \in {\frac{\pi}{2}}$

$y=\frac{1}{3}[\sin x+[\sin x+[\sin x]]]$

$y=1$

$[1+[1]]=2 \cos x$

$\cos x=1$ which is possible at $x=0$ hence NO SOLUTION

Similary if $\sin x \in (0,\pi)-\frac{\pi}{2}$

$y=\frac{1}{3}[\sin x+[\sin x+[\sin x]]]$

$y=0$

$[0+[0]]=2 \cos x$

$\cos x=0$ which is possible at $x=\frac{\pi}{2}$ hence NO SOLUTION

My answer is zero but the official answer is 3 "three".

Please help me understand my mistake.

Answer 1

I believe that your answer is correct. $[x+k]=[x]+k$ if $k$ is an integer, so the given pair of equations is just equivalent to $y=[\sin x]$ and $[y]=\cos x$.

This gives $\cos x=[[\sin x]]=[\sin x]$ which has no solution.