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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

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    $\begingroup$ $\cos x - 2$ is always negative $\endgroup$ – Azlif Nov 23 '19 at 5:09
  • $\begingroup$ Be careful when you multiply through by cos x. If $cos x < 0$ it is going to flip the sign of the inequality. $\endgroup$ – Doug M Nov 23 '19 at 5:21
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The inequality should be factorized as,

$$\frac{\sqrt{2}}{\cos x}\cos\left(x+\frac{\pi}{4} \right )\left(\cos x -2 \right )\le0$$

Two cases:

1) $\cos x > 0, \>\>\>\> \cos\left(x+\frac{\pi}{4} \right )\ge 0$, which leads to $$-\frac\pi2 +2\pi k < x \le \frac\pi4+2\pi k$$

2) $\cos x < 0, \>\>\>\> \cos\left(x+\frac{\pi}{4} \right )\le 0$, which leads to $$\frac{\pi}2 +2\pi k < x \le \frac{5\pi}4+2\pi k$$

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Better is to write $$(\cos(x)-2)(\tan(x)-1)\geq 0$$

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$\cos x + 2 \tan x < 2 + \sin x$

$\cos x + 2 \frac {\sin x}{\cos x} - \sin x - 2 < 0$

$\frac {\cos^2 x + 2\sin x - \sin x\cos x - 2\cos x}{\cos x} < 0\\ \frac {\cos x (\cos x - 2) + \sin x(2-\cos x)}{\cos x} < 0\\ \frac {(\cos x-\sin x) (\cos x - 2)}{\cos x} < 0$

$\cos x - 2 < 0$

$\frac {(\cos x-\sin x)}{\cos x} > 0$

$\cos x > \sin x$ 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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  • $\begingroup$ Finally the complete solution $\pmod {2\pi}$. (+1) $\endgroup$ – trancelocation Nov 23 '19 at 6:08

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