<s>[INST] 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
$$ 2x \leq  2\pi + (\pi - x)$$ 
$$ x \leq \pi$$
 [/INST] $$0\le\cos x+\cos2x=2\cos\dfrac{3x}2\cos\dfrac x2=2\cos^2\dfrac x2\left(4\cos^2\dfrac x2-3\right)$$
$$\implies0\le4\cos^2\dfrac x2-3=2(1+\cos x)-3=2\cos x-1\iff\cos x\ge\dfrac12$$
$2n\pi-\dfrac\pi3\le x\le2n\pi+\dfrac\pi3$ where $n$ is any integer
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