Verified answer

cos²Θ - 3cosΘ = 3

let cosΘ = x

x² - 3x = 3

x² - 3x - 3 = 0

Quadratic formula:

for ax² + bx +c = 0

x = [-b +/- sqrt(b² - 4ac)] / (2a)

x = [-(-3) +/- sqrt((-3)² - 4(-3))] / (2)(1)

x = [3 +/- sqrt(9 + 12] / (2)

x = [3 +/- sqrt(21)] / (2)

so now we can plug in x = cosΘ.

cosΘ = [3 +/- sqrt(21)] / (2)

Θ = arccos [ (3+sqrt(21))/2 ] and

Θ = arccos [ (3-sqrt(21))/2 ]

but arccos (cos^(-1)) is only defined from -1< x < 1.

( 3/2 + sqrt(21)/2 ) > 1 so this answer can be thrown out.

Θ = arccos [ (3-sqrt(21))/2 ]

plugging this into a calculator

Θ = 142.3 degrees