# 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

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

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 $$\frac{\pi}2 +2\pi k < x \le \frac{5\pi}4+2\pi k$$

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

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

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

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

$$\cos x - 2 < 0$$

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

$$\cos x > \sin x$$ and $$\cos x > 0$$ or $$\cos x < \sin x$$ and $$\cos x < 0$$

Over the interval $$[0,2\pi)$$

$$[0 , \frac {\pi}{4}) \cup (\frac {\pi}{2},\frac {5\pi}{4})\cup (\frac {3\pi}{2}, 2\pi)$$

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