<s>[INST] Solve $y=\frac{1}{3}[\sin x+[\sin x+[\sin x]]]$ & $[y+[y]]=2\cos x$ $[x]$ 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.
 [/INST] 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.
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