Question

Тригонометрические уравнения

Answer 1

Ответ:

$a)\ \ sin(2x+\dfrac{\pi}{4})=-\dfrac{\sqrt3}{2}\\\\2x+\dfrac{\pi}{4}=(-1)^{n}\cdot (-\dfrac{\pi}{3})+\pi n\ \ ,\ \ 2x=-\dfrac{\pi}{4}+(-1)^{n+1}\cdot \dfrac{\pi}{3}+\pi n\ ,\\\\x=-\dfrac{\pi}{8}+(-1)^{n+1}\cdot \dfrac{\pi}{6}+\dfrac{\pi n }{2}\ ,\ n\in Z$

$b)\ \ cos(\dfrac{x}{4}-\dfrac{\pi}{3})=\dfrac{1}{2}\\\\\dfrac{x}{4}-\dfrac{\pi}{3}=\pm \dfrac{\pi}{3}+2\pi n\ \ ,\ \ \dfrac{x}{4}=\dfrac{\pi}{3}\pm \dfrac{\pi}{3}+2\pi n\ \ ,\\\\x=\dfrac{4\pi}{3}=\pm \dfrac{4\pi}{3}+8\pi n\ \ ,\ \ n\in Z$

$c)\ \ tg(2x+\dfrac{\pi}{3})=-\dfrac{1}{\sqrt3}\\\\2x+\dfrac{\pi}{3}=-\dfrac{\pi}{6}+\pi n\ \ ,\ \ \ 2x=-\dfrac{\pi}{3}-\dfrac{\pi}{6}+\pi n\ \ ,\ \ 2x=-\dfrac{\pi}{2}+\pi n\ \ ,\\\\x=-\dfrac{\pi}{4}+\dfrac{\pi n}{2}\ ,\ \ n\in Z$

$d)\ \ cos4x=-\dfrac{8}{3}<-1\ \ \ \Rightarrow \ \ \ \ x\in \varnothing$