1) Решите неравенство:2) Решите уравнения:

Решите задачу:

$\frac{3x^2+4x}{9-x} \geq 0\; \; \to \; \; \frac{x(3x+4)}{-(x-9)} \geq 0\; \; \to \; \; \frac{x(3x+4)}{x-9}\leq 0\\\\---[\frac{-4}{3}]+++[0]---(9)+++\\\\x\in (-\infty,-\frac{4}{3}]U[0,9)$

$2)\; cos\frac{x}{2}=\frac{\sqrt2}{2}\\\\\frac{x}{2}=\pm \frac{\pi}{4}+2\pi n\\\\x=\pm \frac{\pi}{2}+4\pi n,n\in Z\\\\3)4sinxcosx=sin^2x+cos^2x\\\\2sin2x=1\\\\sin2x=\frac{1}{2}\\\\2x=(-1)^{n}\frac{\pi}{6}+\pi n\\\\x=(-1)^{n}\frac{\pi}{12}+\frac{\pi n}{2},n\in Z$