The number of integral values of k belonging to  $\left[0,2\right]\cup \left[5,\infty \right]$  such that the equation  $k\cos x-3\sin x=k+1$ possess a solution is

$k\cos x-3\sin x=k+1\Rightarrow$$\frac{k}{\sqrt{k^2+9}}\cos x-\frac{3}{\sqrt{k^2+9}}\sin x=\frac{k+1}{\sqrt{k^2+9}}$

$\Rightarrow \cos \left(x+\varphi \right)=\frac{k+1}{\sqrt{k^2+9}}, where \tan \varphi =\frac{3}{k}$

$\Rightarrow -1\le \frac{k+1}{\sqrt{k^2+9}}\le 1 \left(\because -1\le \cos \left(x+\varphi \right)\le 1\right)$

$$\begin{gathered} \Rightarrow {\left(k+1\right)}^2\le k^2+9 \\ \Rightarrow 2k+1\le 9 \\ \Rightarrow k\le 4 \\ \therefore In \left[0,2\right]\cup \left[5,\infty \right], k takes the integer values 0,1,2 only. \end{gathered}$$