Question

Из c1 решить ) cos2x-sin²x=0,25 при x ∈ [π/2; 2π]

Answer 1

cos2x=1-2sin^2x

1-3sin^2x-1/4=0

3sin^2x=3/4

sinx=+-1/2

x=5/6П

x=7/6П

x=11/6П

Answer 2

$cos2x-sin^2x=0.25\\ cos^2x-sin^2x-sin^2x=0.25\\ 1-sin^2x-sin^2-sin^2x=0.25\\ 1-3sin^2x=0.25\\ -3sin^2x=0.25-1\\ -3sin^2x=-0.75\\ sin^2x=0.25\\ sinx= \frac{1}{2}\\ x_{1}=\frac{\pi}{6}+2\pi n\\ x_{2}=\frac{5\pi}{6}+2\pi n\\ sin x = -\frac{1}{2}\\ x_{3}=-\frac{\pi}{6}+2\pi n\\ x_{4}=-\frac{5\pi}{6}+2\pi n\\$

$\frac{5\pi}{6};\frac{7\pi}{6};\frac{11\pi}{6}$