Question

Solve the following equation exactly on the interval [0,2π).

3cos2θ−2cosθ−2=0
Select all correct answers, assuming they are rounded to two decimal places.

Select all that apply:

θ≈1.24
θ≈2.15
θ≈2.41
θ≈3.55
θ≈4.13
θ≈5.36

Answer 1

Answer:θ≈2.15

θ≈4.13 are correct

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

3u^2 - 2u - 2 = 0

We can use the quadratic formula to solve for u:

u = (-b ± sqrt(b^2 - 4ac)) / 2a

where a = 3, b = -2, and c = -2. Substituting these values, we get:

u = (2 ± sqrt(4 + 24)) / 6

u = (2 ± 2sqrt(7)) / 6

Simplifying this expression, we get:

u = (1 ± sqrt(7)) / 3

Therefore, either: