Question

Задание из C1 помогите решить пожалуйста) 

cos2x-sin²x=0,25 при x ∈ [π/2; 2π]

Answer 1

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

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

Answer 2

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