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. 
 A: 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 $\ge0\implies(y-2)(y+2)\le0\iff\cdots$
A: My Solution::  Given $$f(\theta) = \cos \theta \left(\sin \theta + \sqrt{\sin^2 \theta + 3}\right)$$
Now let $$y=\sin \theta \cdot \cos \theta  +\cos \theta \cdot \sqrt{\sin^2 \theta + 3}$$
Now using the Cauchy-Schwarz inequality, we get
$$\left(\sin^2 \theta +\cos^2 \theta \right)\cdot \left\{\cos^2 \theta + \left(\sqrt{\sin^2 \theta + 3}\right)^2\right\}\geq \left\{\sin \theta \cdot \cos \theta +\cos \theta \cdot \sqrt{\sin^2 \theta + 3}\right\}^2$$
So we get $$y^2 \leq \left(\sin^2 \theta +\cos^2 \theta \right)\cdot \left\{\cos^2 \theta + \sin^2 \theta + 3\right\}=2^2$$
So we get $$-2 \leq y\leq 2\Rightarrow y\in \left[-2,2\right]$$
A: 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
$$-2\le\cos{\theta}(\sin{\theta}+\sqrt{\sin^2{\theta}+3})\le 2$$
