2 кос^2 х - кос х - 1=0

2cos^2x-cosx-1=0

cosx=t, t[-1; 1]

2t^2-t-1=0

Д=1+4*2=9

t1 = (1+3) / 4=1

t2 = (1-3) / 4=-2/4=-0,5

cosx=1

x=2 пк; кэz

cosx=-0,5

x=±2 п/3+2 пк; 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.