Question

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

3cos2(θ)+7cos(θ)−8=0
Select all that apply:

0.3
0.74
0.57
1.98
1.65
5.71

Answer 1

Answer:correct answer is 0.57

5.71

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

3u^2 + 7u - 8 = 0

Now we can factor the quadratic:

(3u - 1)(u + 8) = 0

Therefore, either:

3cos(θ) - 1 = 0

cos(θ) = 1/3

or:

cos(θ) + 8 = 0

cos(θ) = -8 (not possible, since cos(θ) is between -1 and 1)

So, we have cos(θ) = 1/3. Since 0 ≤ θ < 2π, we can find the two solutions in the interval [0, 2π) by using the inverse cosine function:

θ = arccos(1/3)

Using a calculator, we find:

θ ≈ 1.2309 or θ ≈ 5.1102

Therefore, in radians, the solutions for θ are approximately:

1.2309 (which is less than 2π)

5.1102 (which is greater than 2π)

So the only answer that satisfies 0 ≤ θ < 2π is:

1.2309 (rounded to four decimal places)

Therefore, the answer is: