Question

Решите уравнение
${sin}^{2}2x = \frac{1}{2}$
​

Answer 1

$sin2x=-\frac{\sqrt{2} }{2}\\\\ 2x=(-1)^{k}(-\frac{\pi}{4}) +\pi k, k \in Z\\\\x=(-1)^{k}(-\frac{\pi}{8}) +\frac{\pi}{2} k, k \in Z$

или

$sin2x=\frac{\sqrt{2} }{2}\\\\ 2x=(-1)^{n}(\frac{\pi}{4}) +\pi n, n \in Z\\\\x=(-1)^{n}(\frac{\pi}{8}) +\frac{\pi}{2} n, n \in Z$

О т в е т.$(-1)^{n}(-\frac{\pi}{8}) +\frac{\pi}{2} k, k \in Z(-1)^{n}(\frac{\pi}{8}) +\frac{\pi}{2} n, n \in Z$

Но лучше так:

$sin2x=-\frac{\sqrt{2} }{2}\\\\ 2x=-\frac{\pi}{4} +2\pi k, k \in Z; 2x=-\frac{3\pi}{4} +2\pi k, k \in Z;\\\\x=-\frac{\pi}{8} +\pi k, k \in Z; 2x=-\frac{3\pi}{8} +\pi k, k \in Z;$

$sin2x=\frac{\sqrt{2} }{2}\\\\ 2x=\frac{\pi}{4} +2\pi n, n \in Z;2x=\frac{3\pi}{4} +2\pi n, n \in Z;\\\\x=\frac{\pi}{8} +\pi n, n \in Z;x=\frac{3\pi}{8} +\pi n, n \in Z;$

О т в е т. $\pm\frac{\pi}{8} +\pi k, k \in Z;\pm\frac{3\pi}{8} +\pi n, n \in Z;$