Question

Решите уравнения
2 cos x/3= -1
2sin2x- корень из 2= 0

Answer 1

1)

$2\cos \dfrac{x}{3} = -1\\\\\\\cos\dfrac{x}{3} = -\dfrac{1}{2}\\\\\\\left[\begin{gathered}\dfrac{x}{3} = \dfrac{2\pi}{3} + 2\pi k\\\\\dfrac{x}{3} = -\dfrac{2\pi}{3} + 2\pi k\\\end{gathered}\ \ \ \Leftrightarrow\ \left[\begin{gathered}x = 2\pi + 6\pi k\\\\x = -2\pi + 6\pi k\\\end{gathered}\ \ \ \ ,\ k\in\mathbb{Z}$

2)

$2\sin 2x - \sqrt{2} = 0\\\\2\sin 2x = \sqrt{2}\\\\\sin 2x = \dfrac{\sqrt{2}}{2}\\\\\\\left[\begin{gathered}2x = \frac{\pi}{4} + 2\pi k\\\\2x = \frac{3\pi}{4} + 2\pi k\\\end{gathered}\ \ \ \Leftrightarrow\ \left[\begin{gathered}x = \frac{\pi}{8} + \pi k\\\\x = \frac{3\pi}{8} + \pi k\\\end{gathered}\ \ \ \ ,\ k\in\mathbb{Z}$