Question

cos2x+sin^2х=0,25
Помогите прошууууууу

Answer 1

$cos2x+sin^2x=0.25 \\ cos^2x-sin^2x+sin^2x= \frac{1}{4}\\ cos^2x= \frac{1}{4}\\ cosx= ^+_- \frac{1}{2}\\ x_{1,2}= ^+_- \frac{\pi }{3}+2\pi n, n \to Z \\ x_{3,4}= ^+_-(\pi - \frac{\pi }{3})+2\pi n = ^+_-\frac{2\pi }{3}+2\pi n, n \to Z$

Answer 2

cos2x+sin²x=0,25
cos²x - sin²x + sin²x = 0,25
cos²x=0,25


cosx=1/2
cosx= -1/2

x₁,₂= +- arccos1/2 + 2πn
x₃,₄= +- arccos (-1/2) + 2πn


x₁,₂ = +- π/3+2πn,n∈z
x₃,₄= +- (π-arccos1/2) + 2πn = +- (π-π/3)+2πn = +- 2π/3 + 2πn, n∈z