Question

Помогите срочно с тригонометрическими уравнениями!!! Даю 100 баллов.

Answer 1

$1) 5\cos^2\frac{2x}{3}-10 \cos\frac{2x}{3} =0|:5\\\\\cos^2\frac{2x}{3}-2\cos\frac{2x}{3}=0;\\\\\cos\frac{2x}{3}(\cos\frac{2x}{3}-2)=0\Rightarrow \cos\frac{2x}{3}=0\\\\\frac{2x}{3}=\frac{\pi}{2}+ \pi n,n\in \mathbb Z|\cdot\frac{3}{2} \\\\x=\frac{3\pi}{4}+\frac{3\pi n}{2},n\in \mathbb Z.$

$\\\\2)\sin(4\pi+\frac{5x}{4})=\sin(\frac{5\pi}{2}-\frac{5x}{4});\\\\ \sin\frac{5x}{4}=\cos \frac{5x}{4} |:\cos \frac{5x}{4}\neq 0\\\\\frac{\sin\frac{5x}{4}}{\cos \frac{5x}{4}} =1,\\\\tg\frac{5x}{4}} =1\\\\\frac{5x}{4}}=arctg(1)+\pi n,n\in\mathbb Z;\\\\\frac{5x}{4}}=\frac{\pi}{4}+ \pi n,n\in\mathbb Z|\cdot\frac{4}{5} \\\\x=\frac{\pi}{5} +\frac{4\pi n}{5}, n\in\mathbb Z$

$3)tg^2(2\pi-3x)+2ctg(\frac{\pi}{2}+3x)-3=0;\\\\ (-tg3x)^2-2tg3x-3=0;\\\\tg^23x-2tg3x-3=0\Rightarrow\\\\\Rightarrow\left \ [ {{tg3x=-1,} \atop {tg3x=3;}} \right. \left \ [ {{3x=arctg(-1)+\pi n, n\in\mathbb Z} \atop {3x=arctg(3)+\pi n, n\in\mathbb Z|:3}} \right. \\\\\left \ [{{3x=-\frac{\pi}{4}+\pi n, n\in\mathbb Z |:3} \atop {x=\frac{1}{3}arctg3+\frac{\pi n}{3},n\in\mathbb Z }} \right.$

$\left \ [ {{x=-\frac{\pi}{12}+\frac{\pi n}{3},n\in\mathbb Z } \atop {x=\frac{1}{3}arctg3+\frac{\pi n}{3},n\in\mathbb Z} \right.$