2кос^2 х - кос х - 1=0

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t1 = (1 + 3)/2*2

t2 = (1 - 3)/2*2

2cos^2x-cosx-1=0cosx=t, t[-1;1]2t^2-t-1=0Д=1+4*2=9t1=(1+3)/4=1t2=(1-3)/4=-2/4=-0,5cosx=1x=2пк;кэzcosx=-0,5x=±2п/3+2пк;кэz

2cos²x - cosx - 1 = 0

Пусть cosx = t, -1 ≤ t ≤ 1

Тогда: 2t² - t - 1 = 0

D = b² - 4ac = 1 - 4*2*(-1) = 1 + 8 = 9

t₁ = (1 + 3)/2*2 = 4/4 = 1

t₂ = (1 - 3)/2*2 = -2/4 = -1/2

1)cosx = 1

x = 2πn, n ∈ Z

2) cosx = -1/2

x = +- arccos(-1/2) + 2πn, n ∈ Z

x = +- (π - arccos1/2) + 2πn, n ∈ Z

x = +- (π - π/3) + 2πn, n ∈ Z

x = +- 2π/3 + 2πn, n ∈ Z

Ответ: x = 2πn, x = +- 2π/3 + 2πn, n ∈ Z.