Question

Полностью решите данные тригонометрические уравнения : 1) 2) 3) 4) 5) 6)

Answer 1

$sin(x-\frac{\pi}{3})+1=0\\sin(x-\frac{\pi}{3})=-1\\x-\frac{\pi}{3}=-\frac{\pi}{2}+2\pi n\\x=-\frac{\pi}{6}+2\pi n, \; n\in Z;\\\\1-cos^22x=\frac{\sqrt2}{2}\\sin2x=\frac{\sqrt2}{2}\\2x=arcsin\frac{\sqrt2}{2}+2\pi n\\x=\frac{\pi}{4}*\frac{1}{2}+\pi n\\x=\frac{\pi}{8}+\pi n, \; n\in Z$

$\sqrt3tg(\frac{\pi}{6}-x)=-1\\tg(x-\frac{\pi}{6})=\frac{\sqrt3}{3}\\x-\frac{\pi}{6}=arctg \frac{\sqrt3}{3}+\pi n\\ x=\frac{\pi}{6}+\frac{\pi}{6}+\pi n\\x=\frac{\pi}{3}+\pi n, \; n\in Z;\\\\ cos(x+\frac{\pi}{6})-1=0\\cos(x+\frac{\pi}{6})=1\\x+\frac{\pi}{6}=\pi+2\pi n\\x=\frac{5\pi}{6}+2\pi n, \; n\in Z;$

$sin\frac{x}{4}cos\frac{x}{4}=-\frac{1}{4}|*2\\2sin\frac{x}{4}cos\frac{x}{4}=-\frac{1}{2}\\sin(2*\frac{x}{4})=-\frac{1}{2}\\\frac{x}{2}=-arcsin\frac{1}{2}+2\pi n\\x=-\frac{\pi}{6}*2+4\pi n\\x=-\frac{\pi}{3}+4\pi n, \; n\in Z;\\\\ \sqrt3ctg(\frac{\pi}{3}-x)=-3\\ctg(x-\frac{\pi}{3})=\sqrt3\\x-\frac{\pi}{3}=\frac{\pi}{6}+\pi n\\ x=\frac{\pi}{2}+\pi n, \; n\in Z$