Find all possible values of \(\cos(\theta)\) if \(\cos(2\theta) = 2\cos(\theta) .\)
+128475 cos (2θ) = 2cos(θ)
cosθ*cosθ - sinθ^sinθ = 2cos(θ)
cos^2θ - 2cosθ - sin^2θ = 0
cos^2θ - 2cosθ - (1 - cos^2θ) = 0
2cos^2θ - 2cosθ - 1 = 0
Let cosθ = x ....and we have that....
2x^2 - 2x - 1 = 0
Using the quadratic formula x =
2 ±√[ 4 + 8]
_________ =
4
2 ±√12
_______ =
4
2 ± 2√3
________ =
4
1 ±√3
_____
2
Only 1 - √3
_______ is a possible solution
2
So
cos θ = 1 - √3
______
2
arccos 1 - √3
_______ = θ ≈ 111.5° + 2pi * n or 258.5° + 2pi * n
2
Where n is an integer
