Question

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 1

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

cos² θ + sin θ + 1 = 0 

⇒ (1 – sin² θ) + sin θ + 1 = 0 

⇒ sin² θ – 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 = \(\frac{3π}{2}\) ∈ \(\big( \frac{5π}{4},\frac{7π}{4}\big)\)