Prove: $\sin 2 \theta \;\ge\; \frac{1}{2}\left(-1-3 \cos^2 \theta\right)$ Please help me prove the following inequality: 

$$\sin 2 \theta \;\ge\; \frac{1}{2}\left(-1-3 \cos^2 \theta\right)$$

I have been working on it for an hour in vain. I have derived $2\sin \theta \cos \theta$ from the left side and $\frac{1}{2}\left(-4+3 \sin^2 \theta\right)$ from the right side, but I don't know where to go from there.

Let $y=2\sin2x+3\cos^2x=\cos^2x(4\tan x+3)$
$$\iff y\tan^2 x-4\tan x+y-3=0$$ which is a Quadratic  Equation in $\tan x$
As $\tan x$ is real, the discriminant must be $\ge0$
i.e., $$4^2-4y(y-3)=-4(y+1)(y-4)\ge0\iff(y+1)(y-4)\le0$$
$$\iff-1\le y\le4$$ 
