Question

решить уравнение
2-5cosx-cos2x=0

Answer 1

2 - 5Cosx - Cos2x = 0
2 - 5Cosx - (2Cos²x - 1) = 0
2 - 5Cosx - 2Cos²x + 1 = 0
2Cos²x + 5Cosx - 3 = 0
Cosx = 1/2                     Cosx = - 3 - решений нет, так как |- 3| > 1
x = + - π/3 + 2πn, n ∈ z

Answer 2

$2-5cosx-cos2x=0 \ 2-5cosx-cos^2x+sin^2x=0 \ 2-5cosx-cos^2x+1-cos^2x=0 \ 2cos^2x+5cosx-3=0 \ \ cosx=t; t in [-1;1]\ \ 2t^2+5t-3=0 \ D=25+24=7^2 \ t_1= dfrac{-5+7}{4}=0,5 \ t_2= dfrac{-5-7}{4}=-3 notin ODZ \ \ cosx=0,5 \ x=б dfrac{ pi }{3}+2 pi k;k in Z$