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. 

Use this well known inequality
$$-\dfrac{a^2+b^2}{2}\le ab\le\dfrac{a^2+b^2}{2},a,b\in R$$
so 
$$-\dfrac{\cos^2{\theta}+4\sin^2{\theta}}{2}\le\cos{\theta}\cdot 2\sin{\theta}\le\dfrac{\cos^2{\theta}+4\sin^2{\theta}}{2}\tag{1}$$
$$-\dfrac{4\cos^2{\theta}+\sin^2{\theta}+3}{2}\le2\cos{\theta}\cdot\sqrt{\sin^2{\theta}+3}\le \dfrac{4\cos^2{\theta}+\sin^2{\theta}+3}{2}\tag{2}$$
then (1)+(2)
$$-4\le2\cos{\theta}(\sin{\theta}+\sqrt{\sin^2{\theta}+3})\le 4$$
so
$$-2\le\cos{\theta}(\sin{\theta}+\sqrt{\sin^2{\theta}+3})\le 2$$
