IOE, question

Rewrite the given equation as

$\sqrt{k^{2} + 9}\left(\dfrac{k}{\sqrt{k^{2} + 9}} \cos \theta - \dfrac{3}{\sqrt{k^{2} + 9}} \sin \theta\right) = k + 1 \Longrightarrow$

$\sqrt{k^{2} + 9} \sin(\alpha - \theta) = k + 1 \Longrightarrow \sin(\alpha - \theta) = \dfrac{k + 1}{\sqrt{k^{2} + 9}}$ , where $\alpha = \sin^{-1}\left(\dfrac{k}{\sqrt{k^{2} + 9}}\right)$ .

As the range of the sine function is $[-1,1]$ , we then require that

$-1 \le \dfrac{k + 1}{\sqrt{k^{2} + 9}} \le 1 \Longrightarrow -\sqrt{k^{2} + 9} \le k + 1 \le \sqrt{k^{2} + 9}$ .

For $k \ge -1$ we have that $k + 1 \le \sqrt{k^{2} + 9} \Longrightarrow (k + 1)^{2} \le k^{2} + 9 \Longrightarrow 2k \le 8 \Longrightarrow k \le 4$ .

For $k \le - 1$ we have that $-\sqrt{k^{2} + 9} \le k + 1 \Longrightarrow k^{2} + 9 \ge (k + 1)^{2} \Longrightarrow 8 \ge 2k \Longrightarrow 4 \ge k$ .

So in general the original equation will have a solution for $\boxed{k \le 4}$ .