Question

A, sin4x = 2/3
b, sinx = 1/4
c, 3sin4x = 5
d, 2sin2x - 1 = 0
e, sin2x = -1

Answer 1

Пошаговое объяснение:

$1. \\sin(4x) = \frac{2}{3} \\4x = (-1)^narcsin(\frac{2}{3}) + \pi n\\ x = \frac{1}{4} ((-1)^narcsin(\frac{2}{3}) + \pi n)\\2.\\sin(x) = \frac{1}{4}\\ x = (-1)^narcsin(\frac{1}{4}) + \pi n\\3.\\3sin(4x) = 5\\sin(4x) = \frac{5}{3} \\No - real - solutions\\4. \\2sin(2x) - 1 = 0\\sin(2x) = \frac{1}{2}\\ 2x = (-1)^narcsin(\frac{1}{2}) + \pi n = (-1)^n\frac{\pi}{6} + \pi n\\x = (-1)^n \frac{\pi}{12} + \frac{\pi}{2} n\\5.\\sin(2x) = -1\\2x = -\frac{\pi}{2} + 2\pi n\\x = -\frac{\pi}{4} +\pi n$