Question

Solve the following for θ, in radians, where 0≤θ<2π.

2cos2(θ)−8cos(θ)−1=0
Select all that apply:

1.07
1.69
4.59
2.49
2
2.33

Answer 1

Answer:correct answers 1.69

4.59

Explanation:

We can solve this quadratic equation in cos(θ) by using the substitution u = cos(θ):

2u^2 - 8u - 1 = 0

Now we can use the quadratic formula to solve for u:

u = (-b ± sqrt(b^2 - 4ac)) / 2a

where a = 2, b = -8, and c = -1. Substituting these values, we get:

u = (8 ± sqrt(8^2 - 4(2)(-1))) / 2(2)

u = (8 ± sqrt(72)) / 4

u = 2 ± sqrt(18)

Therefore, either: