Find all solutions of the given equation.

Question

Round terms to two decimal places where appropriate.) 
4 cos2 θ − 1 = 0

Answer 1

General form:

4cos2θ - 1 = 0

4cos2θ = 1

cos2θ  = 1/4

2θ = cos^-1(1/4)

Let us assume that 2θ = t.

The genaral solution of cos(x) = k is x = 2nπ ± α where α = cos^-1(k), where n is an integer. 

t = 2nπ ± cos^-1(1/4) where n Є z.

2θ = 2nπ ± cos^-1(1/4)

θ = 1/2  (2nπ ± cos^-1(1/4))

θ = (2nπ ± cos^-1(1/4)) / 2

θ = (2nπ ± 75.5224) / 2

θ = (2nπ ± 75.52) / 2                   [Rounded to two decimals]

θ = (2nπ + 75.52) / 2 or (2nπ - 75.52) / 2 where n Є z.

Comments

The genaral solution (in radians) of cos(x) = k is x = 2nπ ± α where α = cos^-1(k), where n is an integer.

The genaral solution (in degrees) of cos(x) = k is x = 360n ± α where α = cos^-1(k), where n is an integer.

The equation is 4 cos(2θ) − 1 = 0.

Solution in radians:

θ = 1/2  ( 2nπ ± cos^-1(1/4) )

θ = nπ ± 1/2 cos^-1(1/4)

θ = nπ ± 1/2 cos^-1(0.25)

θ ≅ 3.14 n ± 1/2 (1.32)                    [Rounded to two decimals]

θ ≅ 3.14 n ± 0.66, where n ∈ Z.