# Trigonometric inequality

I'm trying to solve the following inequality, but I can't seem to be able to factor it:

$$5\sin^2{}x>\cos{x}(3\sin{x}+2\cos{x})$$ $$5\sin^2x-3\cos{x}\sin{x}-2\cos^2{x}>0$$ $$5(1-\cos^2{x})-3\cos{x}\sin{x}-2\cos^2{x}>0$$

Any hints on how can I solve it?

• Should be $5\sin x$, not $5\sin^2x$ in second line. Mar 27, 2016 at 17:51
• $5sin^2x+2sinxcosx-5sinxcosx-2cos^2x$. Now factor by grouping Mar 27, 2016 at 17:52
• @bgins Sorry, that was a typo. Fixed now Mar 27, 2016 at 17:53

Just to see a different approach. we can write the inequality as:

$$5\sin^2 x-5\sin x \cos x+2 \sin x \cos x-2 \cos^2 x >0$$ $$(\sin x- \cos x)(5 \sin x+2 \cos x)>0$$ now let $\delta$ such that: $$\cos \delta= \frac {5}{\sqrt{5^2+2^2}}\qquad \sin \delta= \frac {2}{\sqrt{5^2+2^2}}$$ the inequality becomes: $$\sqrt{29}(\sin x - \cos x)(\cos \delta \sin x+\sin \delta \cos x)>0$$

$$(\sin x - \cos x)\sin (\delta+x)>0$$

that you can solve with simple trigonometry.

1) If $\cos x=0$ then $5 \sin^2 x>0$

2) If $\cos x \not =0$ then $$5\tan^2x - 3 \tan x-2>0$$ $\tan x<-\frac 25$ or $\tan x>1$

$x<-\arctan \left(\frac 25 \right)+\pi n$ or $x>\frac{\pi}4+\pi k$ or $x=\frac{\pi}2+\pi m, n.k.m \in \mathbb Z$

HINT :

Case 1 : Consider when $\cos x=0$.

Case 2 : When $\cos x\not=0$, divide the both sides of $$5\sin^2 x-3\cos x\sin x-2\cos^2x\gt 0$$ by $\cos^2x\gt 0$ to get $$5\tan^2 x-3\tan x-2\gt 0$$

$$5\sin^2 x>\cos x(3\sin x+2\cos x)$$ $$5\sin^2x-3\cos x\sin x-2\cos^2 x>0$$ $$(5\sin x+2\cos x)(\sin x-\cos x)>0$$ $$\left(\tan x+\tfrac25\right)(\tan x-1)>0 \qquad\text{or}\qquad \left(\cot x+\tfrac52\right)(\cot x-1)<0$$ $$\tan x\not\in\left[-\tfrac25,1\right] \qquad\text{or}\qquad \cot x\in\left(-\tfrac52,1\right)$$ $$x+k\pi\in\left(\tfrac\pi4,\pi-\tan^{-1}\tfrac25\right) \qquad\text{for}\quad k\in\mathbb{Z}$$ The solution has period $\pi$ because the locus includes angles $x$ on the unit circle lying strictly between the lines (above or below both) $v=u$ and $v=-\tfrac25u$ in the $uv$ plane.