How many real solutions are there to the equation 2cos4θ − 5cos2θ + 3 = 0 in the interval 0 ≤ θ ≤ 2π?
- A1
- B2
- C3
- D4
- E5
- F6
- G7
- H8
Show the answer and worked solution
answer · C
- A1
- B2
- C3
- D4
- E5
- F6
- G7
- H8
Put u = cos2θ, so 2u2 − 5u + 3 = 0 and (2u−3)(u−1) = 0. The root u = 32 is impossible because cos2θ ≤ 1, leaving cos2θ = 1, that is cosθ = ± 1. In [0, 2π] that gives θ = 0, π and 2π — three solutions.