<s>[INST] Solve inequality $\cos(x)+2\tan(x)\le2+\sin(x)$ Solve inequality 

$\cos(x)+2\tan(x)\le2+\sin(x)$

My proof: 
$\cos(x)+2\tan(x)\le2+\sin(x)\\\cos^2(x)+2\sin(x)\le2\cos(x)+\sin(x)\cos(x)\\\cos(x)\left(\cos(x)-2 \right )+\sin(x)\left(2-\cos(x) \right )\le0\\\left(\cos(x)-\sin(x) \right )\left(\cos(x)-2 \right )\le0\\\sqrt{2}\left(\sin\left(x-\frac{\pi}{4} \right ) \right )\left(\cos(x)-2 \right )\le0\\-\sqrt{2}\cos\left(x+\frac{\pi}{4} \right )\left(\cos(x)-2 \right )\le0$
I stopped at this moment and I have no idea what to do now
 [/INST] $\cos x + 2 \tan x < 2 + \sin x$
$\cos x + 2 \frac {\sin x}{\cos x} - \sin x - 2 < 0$
$\frac {\cos^2 x + 2\sin x - \sin x\cos x - 2\cos x}{\cos x} < 0\\
\frac {\cos x (\cos x - 2) + \sin x(2-\cos x)}{\cos x} < 0\\
\frac {(\cos x-\sin x) (\cos x - 2)}{\cos x} < 0$
$\cos x - 2 < 0$
$\frac {(\cos x-\sin x)}{\cos x} > 0$
$\cos x > \sin x$ and $\cos x > 0$
or $\cos x < \sin x$  and $\cos x < 0$
Over the interval $[0,2\pi)$
$[0 , \frac {\pi}{4}) \cup (\frac {\pi}{2},\frac {5\pi}{4})\cup (\frac {3\pi}{2}, 2\pi)$
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