Verified answer

4cos²θ + 4cosθ - 1 = 0

so, using the quadratic formula we get:

cosθ = [-4 ± √32]/8

=> cosθ = [-4 ± 4√2]/8

i.e. cosθ = [-1 ± √2]/2

so, cosθ = 0.21 or -1.21...not possible

=> θ = ±78° + 360n°...for n = 0, 1, 2,...

i.e.θ = 78° or 282° for 0° ≤ θ ≤ 360°

:)>

4cos^2 t + 4cost -1 =0

cos t = (-4 +/- sqrt(16+16))/8 = (-4 +/- 4root2)/8 = (-1+/- root2)/2

cos t = (-1+root2)/2 or (-1-root2)/2

(-1-root2/2) is less than -1 so ignored.

cost = 0.207 so t = 78, 282

Let x = cos(Θ)

4x^2 + 4x - 1 = 0

x = 0.207106781... or x = 1.207106781.

Disregard second option for real angle.

Solve cos(Θ) = 0.207106781..

(you do that bit)

Regards - Ian H

4cos^2A+cosA-1=0 then

cosA = [-4+/-sqrt(4^2-4.4.(-1)]/(2.4)

cosA = [-4+/-sqrt(16+16)]/8

cosA = [-4+/-sqrt(16.2)]/8

cosA = [-4+/-4sqrt(2)]/8

cosA = 4[-1+/-sqrt(2)]/8

cosA =[-1+sqrt(2)]/2 and [-1-sqrt(2)]/2

cosA = [-1+1.414]/2 and [-1-1.414]/2

cosA = [0.414]/2 =0.207 a positive, angle A is acute (find from table)

and

cosA = [-2.414]/2 = -1.207, which negetive, angle A is obtuse (find angle from table)

both value is admissible because given interval is [0,360]