How many values of θ will satisfy the equation (sin2 θ - 4 sin θ + 3) (4 - cos2 θ + 4 sin θ) = 0, where 0 < θ < \(\frac{\pi}{2}\) ?

Given:

(sin2 θ - 4 sin θ + 3) (4 - cos2 θ + 4 sin θ) = 0, 0 < θ < \(\frac{\pi}{2}\)

Calculation:

According to the question

(sin2 θ - 4 sin θ + 3) (4 - cos2 θ + 4 sin θ) = 0

⇒ (sin2 θ - 4 sin θ + 3) = 0 OR (4 - cos2 θ + 4 sin θ) = 0

Solving the first equation:

(sin2 θ - 4 sin θ + 3) = 0

⇒ sin2 θ - sin θ - 3 sin θ + 3 = 0

⇒ sin θ (sin θ - 1) - 3 (sin θ - 1) = 0

⇒ (sin θ - 1) (sin θ - 3) = 0

⇒ sin θ = 1 OR sin θ = 3

⇒ θ = 90⁰ = \(\frac{\pi}{2}\) (∵ -1 < sin θ < 1, neglecting sin θ = 3)

Solving the another equation:

(4 - cos2 θ + 4 sin θ) = 0

⇒ 4 - (1 - sin2 θ) + 4 sin θ = 0

⇒ sin2 θ + 4 sin θ + 3 = 0

⇒ sin2 θ + sin θ + 3 sin θ + 3 = 0

⇒ sin θ (sin θ + 1) + 3 (sin θ + 1) = 0 

⇒ (sin θ + 1) (sin θ + 3) = 0

⇒ sin θ = -1 OR sin θ = -3

⇒ θ = 270⁰ = \(\frac{3\pi}{2}\) (∵ -1 < sin θ < 1, neglecting sin θ = -3)

Again, according to the question 0 < θ < \(\frac{\pi}{2}\)

∴ The given equation does not satisfy for any value of θ, 0 < θ < \(\frac{\pi}{2}\).