Cos(π(2x+9)/3)=½ tg(π(x-5)/3)= -√3 sin(πx/3)=0,5

Решите задачу:

$1)\; \; cos\frac{\pi (2x+9)}{3} = \frac{1}{2} \\\\ \frac{\pi (2x+9)}{3} =\pm \frac{\pi}{3}+2\pi n,\; n\in Z\\\\2x+9=\pm 1+6n\; ,\; n\in Z\\\\2x=\pm 1-9+6n\; ,\; n\in Z\\\\x=\pm 0,5-4,5+3n\; ,\; n\in Z\\\\2)\; \; tg\frac{\pi (x-5)}{3}=-\sqrt3\\\\\frac{\pi(x-5)}{3}=-\frac{\pi}{3}+\pi n,\; n\in Z\\\\x-5=-1+3n\; ,\; n\in Z\\\\x=4+3n\; ,\; n\in Z$

$3)\; \; sin\frac{\pi x}{3}=0,5\\\\\frac{\pi x}{3}=(-1)^{n}\frac{\pi}{6}+\pi n,\; n\in Z\\\\x=(-1)^{n}\cdot \frac{1}{2}+3n\; ,\; n\in Z$