Question

The number of elements in the set S = {θ∈ [–4π, 4π]: 3cos²2θ + 6cos²θ – 10cos²θ + 5 = 0} is ______.

विकल्प

• 29

• 27

• 32

• 34

Answer 1

The number of elements in the set S = {θ∈ [–4π, 4π]: 3cos²2θ + 6cos²θ – 10cos²θ + 5 = 0} is 32.

Explanation:

Given: 3cos²2θ + 6cos²θ – 10cos²θ + 5 = 0; θ ∈ [–4π, 4π]

⇒ 3cos²2θ + 6cos2θ – 5(1 + cos²θ) + 5 = 0  ...{∵ 2cos²θ = 1 + cos2θ}

⇒ 3cos²2θ + cos2θ = 0

⇒ cos2θ(3cos2θ + 1) = 0

⇒ cos2θ = 0 or cos2θ = `-1/3`

Case-1 If cos 2θ = 0

⇒ 2θ = `(2n + 1) π/2; n ∈ I`

⇒ θ = `(2n + 1) π/4`

⇒ θ = `+-π/4, +-3π/4, +-5π/4, ......... +-15π/4`

∴ For θ∈ [–4π, 4π], 16 values of θ is possible for this case.

Case-2 If cos 2θ = `-1/3`

Let cos α = `-1/3`

⇒ α = `cos^-1(-1/3);α∈(π/2, π)`

⇒ 2θ = `2n π +- α; α∈(π/2, π)`

⇒ θ = `n π +- α/2`

∴  For θ∈ [–4π, 4π], 16 values of θ is possible for this case.

So, number of elements in the set S is 32.