The number of real roots of the equation sin4 θ – 2 sin2 θ – 1

1.	The number of real roots of the equation sin4 θ – 2 sin2 θ – 1 = 0 in the interval (0, 2π) is (a) 0 (b) 1 (c) 2 (d) 4
Answer» Answer : (a) 0 sin⁴ θ – 2 sin² θ – 1 = 0 ⇒ (sin² θ)²– 2 sin² θ – 1 = 0 ⇒ sin² θ = \( {2 \,\pm \sqrt{4-4\times 1 \times (-1)} \over 2\times 1}\) \( = {2\, \pm\, \sqrt{8} \over 2}\) = \( {2\, \pm \,2\sqrt{2} \over 2}\) = 1- \( \sqrt{2} \) or 1 + \( \sqrt{2} \) sin²θ = 1 – \( \sqrt{2} \) is inadmissible as it is not real. ∵ –1 ≤ sin θ ≤ 1 ⇒ 0 ≤ sin² θ ≤ 1 ⇒ sin² θ = 1 + \( \sqrt{2} \) is not possible. Hence the given equation has no real root.

The number of real roots of the equation sin4 θ – 2 sin2 θ – 1 = 0 in the interval (0, 2π) is (a) 0 (b) 1 (c) 2 (d) 4

Answer»

Answer : (a) 0

sin⁴ θ – 2 sin² θ – 1 = 0

⇒ (sin² θ)²– 2 sin² θ – 1 = 0

⇒ sin² θ = \( {2 \,\pm \sqrt{4-4\times 1 \times (-1)} \over 2\times 1}\)

\( = {2\, \pm\, \sqrt{8} \over 2}\) = \( {2\, \pm \,2\sqrt{2} \over 2}\)

= 1- \( \sqrt{2} \) or 1 + \( \sqrt{2} \)

sin²θ = 1 – \( \sqrt{2} \) is inadmissible as it is not real.

∵ –1 ≤ sin θ ≤ 1

⇒ 0 ≤ sin² θ ≤ 1

⇒ sin² θ = 1 + \( \sqrt{2} \) is not possible.

Hence the given equation has no real root.