The solution of the equation $\mathrm{kcos}\mathrm{x}-3\sin \mathrm{x}=\mathrm{k}+1$ is possible if

Here, k cos x - 3 sin x = k +1 could be rewritten as

$\begin{array}{l}\frac{\mathrm{k}}{\sqrt{{\mathrm{k}}^2+9}}\cos \mathrm{x}-\frac{3}{\sqrt{{\mathrm{k}}^2+9}}\sin \mathrm{x}=\frac{\mathrm{k}+1}{\sqrt{{\mathrm{k}}^2+9}}\\ \Rightarrow \cos (\mathrm{x}+\mathrm{\varphi })=\frac{\mathrm{k}+1}{\sqrt{{\mathrm{k}}^2+9}}\end{array}$

$\begin{array}{r}\mathrm{where} \cos \mathrm{\varphi }=\frac{\mathrm{k}}{\sqrt{{\mathrm{k}}^2+9}}\\ \mathrm{and} \sin \mathrm{\varphi }=\frac{3}{\sqrt{{\mathrm{k}}^2+9}}\end{array}$

which posses solution only, if

$\begin{array}{l}-1\le \frac{\mathrm{k}+1}{\sqrt{{\mathrm{k}}^2+9}}\le 1\\ \Rightarrow \left|\frac{\mathrm{k}+1}{\sqrt{{\mathrm{k}}^2+9}}\right|\le 1\\ \Rightarrow (\mathrm{k}+1{)}^2\le {\mathrm{k}}^2+9\\ \Rightarrow {\mathrm{k}}^2+2\mathrm{k}+1\le {\mathrm{k}}^2+9\\ \Rightarrow \mathrm{k}\le 4\end{array}$

Hence, the interval in which kcos x - 3sin x = k +1 admits solution for k is $(-\mathrm{\infty },4]$