The solution of equation cos2 θ + sin θ + 1 = 0 lies in the interval

1.	The solution of equation cos2 θ + sin θ + 1 = 0 lies in the interval:(a) \(\big(- \frac{π}{4} , \frac{π}{4}\big)\)(b) \(\big( \frac{π}{4} , \frac{3π}{4}\big)\)(c) \(\big( \frac{3π}{4} , \frac{5π}{4}\big)\)(d) \(\big( \frac{5π}{4} , \frac{7π}{4}\big)\)
Answer» Answer : (d) \(\big( \frac{5π}{4},\frac{7π}{4}\big)\) cos² θ + sin θ + 1 = 0 ⇒ (1 – sin² θ) + sin θ + 1 = 0 ⇒ sin²θ – sin θ – 2 = 0 ⇒ (sin θ + 1) (sin θ – 2) = 0 ⇒ (sin θ + 1) = 0 or (sin θ – 2) = 0 ⇒ sin θ = –1 (∵ sin θ = 2 is inadmissible) ⇒ sin θ = sin \(\frac{3π}{2}\) ⇒ θ = \(\frac{3π}{2}\) ∈ \(\big( \frac{5π}{4},\frac{7π}{4}\big)\)

The solution of equation cos2 θ + sin θ + 1 = 0 lies in the interval:(a) \(\big(- \frac{π}{4} , \frac{π}{4}\big)\)(b) \(\big( \frac{π}{4} , \frac{3π}{4}\big)\)(c) \(\big( \frac{3π}{4} , \frac{5π}{4}\big)\)(d) \(\big( \frac{5π}{4} , \frac{7π}{4}\big)\)

Answer»

Answer : (d) \(\big( \frac{5π}{4},\frac{7π}{4}\big)\)

cos² θ + sin θ + 1 = 0

⇒ (1 – sin² θ) + sin θ + 1 = 0

⇒ sin²θ – sin θ – 2 = 0

⇒ (sin θ + 1) (sin θ – 2) = 0

⇒ (sin θ + 1) = 0 or (sin θ – 2) = 0

⇒ sin θ = –1 (∵ sin θ = 2 is inadmissible)

⇒ sin θ = sin \(\frac{3π}{2}\)

⇒ θ = \(\frac{3π}{2}\) ∈ \(\big( \frac{5π}{4},\frac{7π}{4}\big)\)