Question

Solve the following for θ, in radians, where 0≤θ<2π.
3cos2(θ)+6cos(θ)−4=0

Answer 1

Answer:1.02 5.27 are correct

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

3u^2 + 6u - 4 = 0

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

u = (-b ± sqrt(b^2 - 4ac)) / 2a

where a = 3, b = 6, and c = -4. Substituting these values, we get:

u = (-6 ± sqrt(6^2 - 4(3)(-4))) / 2(3)

u = (-6 ± sqrt(84)) / 6

u = (-3 ± sqrt(21)) / 3

Therefore, either:

Answer 2

0 ≤ < 2

Explanation: