If $f.(0,\pi )\to R$ is given by $f\left(x\right)=\sum _{k=1}^n [1+\sin kx],\left[x\right]$ denotes the greatest integer function, then the range of $f\left(x\right)$ is

$f\left(x\right)=\sum _{k=1}^n 1+[\sin kx]=n+[\sin x]+[\sin 2x]+\cdots$
$+\left[\sin nx\right].$ If $kx\ne \frac{\pi }{2}$ for any $k=1, 2 .........n$ then $0$
$[\sin kx]=0,k=1,2\dots n\text{.}$ i.e. $f\left(x\right)=n\text{.}$ If $kx=\frac{\pi }{2}$
for some $k$ then $x=\frac{\pi }{2k}$, hence $\sin x,\sin 2x,\dots \sin (k-1)x$ will lie between $0$ and $1$ so $\left[\sin jx\right]=0$
$1\le j\le k-1; \sin kx=1$ so $f\left(x\right)$ can be $n+1$ or $n.$