The values of $p$ for which the equation $3{\sin }^{2}x+12\cos x-3=p$ has at least one solution are

we have, 

$\begin{array}{l}3{\sin }^{2}x+12\cos x-3=p\\ \Rightarrow 3{\cos }^{2}x-12\cos x+p=0\\ \Rightarrow \cos x=\frac{12\pm \sqrt{144-12p}}{6}\end{array}$

This will have a solution, if  

$-1\le \frac{12+\sqrt{144-12p}}{6}\le 1$ and $144-12p\ge 0$

$\Rightarrow -6\le 12\pm \sqrt{144-12p}\le 6$ and $p\le 12$

$\Rightarrow -18\le \pm \sqrt{144-12p}\le -6$ and $p\le 12$

$\Rightarrow -18\le -\sqrt{144-12p}\le -6$ and $p\le 12$

$\Rightarrow 6\le \sqrt{144-12p}\le 18$ and $p\le 12$

$\Rightarrow 36\le 144-12p\le 324$and $p\le 12$ 

$\Rightarrow -108\le -12p\le 180$ and $p\le 12$

$\Rightarrow -15\le p\le 9$ and $p\le 12$

$\Rightarrow -15\le p\le 9$