Solve trigonometric inequality $\cos x \geq \sin^2 x - \cos^2 x $ Solve trigonometric inequality $$\cos x  \geq \sin^2 x - \cos^2 x $$
My incorrect solution: 
$$\cos^2 x-\sin^2 x \geq -\cos x $$
$$\cos 2x \geq \cos (\pi - x) $$
which means:
$$ 2x \geq -(\pi + x)$$ 
$$ x \geq -\pi $$ 
Which is wrong.
And
$$ 2x \leq  2\pi + (\pi - x)$$ 
$$ x \leq \pi$$

\begin{align*}
\cos^2x − \sin^2x & \geq −\cos x\\
\cos^2x − \cos^2x-1 & \geq −\cos x\\
2\cos^2x + \cos x + 1 & \geq 0
\end{align*}
solving for cosx
$$\cos x = [-\infty,1] \cup \left[\frac{1}{2},1\right]$$
rejecting 1st part
$$\cos x \geq \frac{1}{2}$$
$$x \leq \frac{\pi}{3}$$
$$x \geq -\frac{\pi}{3}$$
