Решите уравнение cos2x – sin2x = –0,25.
Источник:
Решение:
cos2x – sin2x = –0,25 |^2
(cos2x – sin2x)2 = (–0,25)2
cos22x – 2sin2x·cos2x + sin22x = \frac{1}{16}
1 – 2sin2x·cos2x = \frac{1}{16} |·(–1)
– 1 + 2sin2x·cos2x = –\frac{1}{16}
2sin2x·cos2x = –\frac{1}{16} + 1
sin(2·2x) = \frac{15}{16}
sin(4x) = \frac{15}{16}
4x = (–1)n·arcsin (\frac{15}{16}) + πn, n∈Z |:4
x = \frac{1}{4}·(–1)n·arcsin (\frac{15}{16}) + \frac{πn}{4}, n∈Z
Ответ: \frac{1}{4}·(–1)n·arcsin (\frac{15}{16}) + \frac{πn}{4}, n∈Z.