How can I show this inequality: $-2 \le \cos \theta (\sin \theta +\sqrt{\sin ^2 \theta +3})\le 2$ 
Show that $$-2 \le \cos \theta ~ (\sin \theta +\sqrt{\sin ^2 \theta +3})\le 2$$ for all value of $\theta$.

Trial: I know that $0\le \sin^2 \theta \le1 $. So, I have $\sqrt3 \le \sqrt{\sin ^2 \theta +3} \le 2 $. After that I am unable to solve the problem. 

Let $\cos \theta ~ (\sin \theta +\sqrt{\sin ^2 \theta +3})=y$
$\iff \sin \theta +\sqrt{\sin ^2 \theta +3}=y\sec\theta$
$\iff \sqrt{\sin ^2 \theta +3}=y\sec\theta-\sin \theta$
Squaring we get $\sin ^2 \theta +3=y^2(1+\tan^2\theta)+\sin^2\theta-2y\tan\theta$
$\iff y^2(\tan^2\theta)-2y(\tan\theta)+y^2-3=0$
As $\tan\theta$ is real, the discriminant must be $\ge0\implies(y-2)(y+2)\le0\iff\cdots$
