Question

Решите уравнение 2sin (3x - π/4) = √2 , на на промежутке (0; 2π)

Answer 1

Ответ:

$2sin(3x-\dfrac{\pi}{4})=\sqrt2\\\\sin(3x-\dfrac{\pi}{4})=\dfrac{\sqrt2}{2}\\\\3x-\dfrac{\pi}{4}=(-1)^{n}\cdot \dfrac{\pi}{4}+\pi n\\\\3x=(-1)^{n}\cdot \dfrac{\pi}{4}+\dfrac{\pi}{4}+\pi n\\\\x=(-1)^{n}\cdot \dfrac{\pi }{12}+\dfrac{\pi }{12} +\dfrac{\pi n}{3}\ ,\ n\in Z\\\\x\in (0;2\pi):\ \ x= \dfrac{\pi}{6}\ ,\ \ \dfrac{\pi }{3}\ ,\ \ \dfrac{5\pi}{6}\ ,\ \ \pi \ ,\ \dfrac{3\pi }{2}\ ,\ \dfrac{5\pi }{3}\ .$