Question

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

−7cos2(θ)+4cos(θ)+6=0
Select all that apply:

1.07
3.96
0.31
2.32
1.68
2.43

Answer 1

Answer:correct answers are 3.96

2.32

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

-7u^2 + 4u + 6 = 0

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

u = (-b ± sqrt(b^2 - 4ac)) / 2a

where a = -7, b = 4, and c = 6. Substituting these values, we get:

u = (-4 ± sqrt(4^2 - 4(-7)(6))) / 2(-7)

u = (-4 ± sqrt(136)) / (-14)

u = (2 ± sqrt(34)) / 7

Therefore, either:

cos(θ) = (2 + sqrt(34)) / 7

or:

cos(θ) = (2 - sqrt(34)) / 7

Since 0 ≤ θ < 2π, we can find the two solutions in the interval [0, 2π) by using the inverse cosine function:

θ = arccos((2 + sqrt(34)) / 7)

θ = arccos((2 - sqrt(34)) / 7)

Using a calculator, we find: