Задача 32092 2cos^3x-cos^2x+2cosx-1=0 2pi 7pi/2...

vk247738147

2018-12-18 19:59:06

2cos^3x-cos^2x+2cosx-1=0 2pi 7pi/2

sova

2018-12-18 20:06:55

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сos^2x*(2cosx-1)+(2cosx-1)=0
(2cosx-1)*(cos^2x+1)=0
2cosx-1=0
cosx=1/2
x= ± (π/3) +2πn, n ∈ Z


(π/3) +2π=(7π/3)∈ [2π; 7π/2]