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.
 A: Hint: Since $1=\sin^2(t)+\cos^2(t)$ and $2\sin(2t)=4\sin(t)\cos(t)$,
$$ 2\sin(2t)+1+3\cos^2(t) = \left(\sin(t)+2\cos(t)\right)^2.$$
A: We need to establish $$\sin2\theta\ge-\dfrac{1+3\cos^2\theta}2=-\dfrac{2+3(1+\cos2\theta)}4$$
$$\iff4\sin2\theta+3\cos2\theta\ge-5$$
Now
$4\sin2\theta+3\cos2\theta=5\sin\left(2\theta+\arcsin\dfrac45\right)$
Finally for real $\theta,$ $$-1\le\sin\left(2\theta+\arcsin\dfrac45\right)\le1$$
A: The extreme values of $A\cos x +B\sin x$ are $\pm\sqrt {A^2+B^2}.$ This is obvious and trivial if $A=B=0.$ If $A,B$ are not both $0,$ there exists $y$ such that $\cos y=A/\sqrt {A^2+B^2}$ and $\sin y=B/\sqrt {A^2+B^2},$ so $$|A\cos x +B\sin x| =|(\sqrt {A^2+B^2}\;)(\cos x \cos y+\sin x \sin y)|$$ $$=(\sqrt {A^2+B^2}\;) |\cos (x-y)|\leq \sqrt {A^2+B^2}.$$
Putting $\cos^2 \theta=(1+\cos 2 \theta)/2,$ the inequality in the Q can be re-arranged as $5\geq -4\sin 2\theta -3\cos 2 \theta .$ Which is true because $|-4\sin 2 \theta -3\cos 2 \theta| \leq \sqrt {(-4)^2+(-3)^2}=5.$
A: 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., $$4^2-4y(y-3)=-4(y+1)(y-4)\ge0\iff(y+1)(y-4)\le0$$
$$\iff-1\le y\le4$$ 
