<s>[INST] 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?
 [/INST] 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.
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