Question

задание 132
полное решение​

Answer 1

$1)\; \; sin8x=1\\\\8x=\frac{\pi }{2}+2\pi k\; ,\; k\in Z\\\\x=\frac{\pi}{16}+\frac{\pi k}{4}\; ,\; k\in Z\\\\\\2)\; \; sin\frac{4x}{11}=0\\\\\frac{4x}{11}=\pi n\; ,\; n\in Z\\\\x=\frac{11\pi n}{4}\; ,\; n\in Z\\\\\\3)\; \; sin(7x-\frac{\pi}{6})=-1\\\\7x-\frac{\pi}{6}=-\frac{\pi}{2}+2\pi k\; ,\; \; 7x=\frac{\pi}{6}-\frac{\pi }{2}+2\pi k\; ,\; k\in Z\; ,\\\\7x=-\frac{\pi }{3}+2\pi k\; ,\; \; x=-\frac{\pi}{21}+\frac{2\pi k}{7}\; ,\; k\in Z$

$4)\; \; sin(3x+\frac{\pi}{3})+1=0\\\\sin(3x+\frac{\pi}{3})=-1\\\\3x+\frac{\pi}{3}=-\frac{\pi}{2}+2\pi n\; \; ,\; \; 3x=-\frac{\pi}{3}-\frac{\pi}{2}+2\pi n\; ,\; n\in Z\\\\3x=-\frac{5\pi}{6}+2\pi n\; ,\; \; \; x=-\frac{5\pi }{18}+\frac{2\pi n}{3}\; ,\; n\in Z$