Question

Решите уравнение cosx=1/2,sin x/2=-1,tg(2x-pi/4)=1

Answer 1

$cosx=\frac{1}{2}\\x=\pm arccos\frac{1}{2}+2\pi*n,\ n\in Z\\x=\pm\frac{\pi}{3}+2\pi*n,\ n\in Z$

$sin\frac{x}{2}=-1\\\frac{x}{2}=-\frac{\pi}{2}+2\pi*n,\ n\in Z\\x=-\pi+4\pi*n,\ n\in Z$

$tg(2x-\frac{\pi}{4})=1\\2x-\frac{\pi}{4}=arctg1+\pi*n,\ n\in Z\\2x=\frac{\pi}{4}+\frac{\pi}{4}+\pi*n, n\in Z\\2x=\frac{\pi}{2}+\pi*n,\ n\in Z\\x=\frac{\pi}{4}+\frac{\pi*n}{2},\ n\in Z$