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?
 A: 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.
A: 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
$$5\tan^2 x-3\tan x-2\gt 0$$ 
A: 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 $x=\frac{\pi}2+\pi m, n.k.m \in \mathbb Z$
A: $$5\sin^2 x>\cos x(3\sin x+2\cos x)$$
$$5\sin^2x-3\cos x\sin x-2\cos^2 x>0$$
$$(5\sin x+2\cos x)(\sin x-\cos x)>0$$
$$
\left(\tan x+\tfrac25\right)(\tan x-1)>0
   \qquad\text{or}\qquad
\left(\cot x+\tfrac52\right)(\cot x-1)<0
$$
$$
\tan x\not\in\left[-\tfrac25,1\right]
   \qquad\text{or}\qquad
\cot x\in\left(-\tfrac52,1\right)
$$
$$
x+k\pi\in\left(\tfrac\pi4,\pi-\tan^{-1}\tfrac25\right)
\qquad\text{for}\quad k\in\mathbb{Z}
$$
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.
