Find the number of solutions of the equation tan x + sec x = 2 cos x,

1.	Find the number of solutions of the equation tan x + sec x = 2 cos x, x∈ [0, π].
Answer» tan x + sec x = 2 cos x ⇒ \(\frac{sin\,x}{cos\,x}\) + \(\frac{1}{cos\,x}\) = 2 cos x ⇒ 1 + sin x = 2 cos²x ⇒ 1 + sin x = 2 (1 – sin²x) = 2 – 2 sin²x ⇒ 2 sin²x + sin x – 1 = 0 ⇒ (sin x + 2) (2 sin x – 1) = 0 ⇒ (sin x + 2) = 0 or 2 sin x = 1 ⇒ sin x = –2 or sin x = \(\frac{1}{2}\) Since sin x = – 2 is inadmissible, therefore, sin x = \(\frac{1}{2}\) ⇒ x = 30°, 150°, i.e. x = \(\frac{π}{6}\) ,\(\frac{5π}{6}\). ∴ The number of solutions x ∈[0, π] are 2.

Find the number of solutions of the equation tan x + sec x = 2 cos x, x∈ [0, π].

Answer»

tan x + sec x = 2 cos x

⇒ \(\frac{sin\,x}{cos\,x}\) + \(\frac{1}{cos\,x}\) = 2 cos x

⇒ 1 + sin x = 2 cos²x

⇒ 1 + sin x = 2 (1 – sin²x)

= 2 – 2 sin²x

⇒ 2 sin²x + sin x – 1 = 0

⇒ (sin x + 2) (2 sin x – 1) = 0

⇒ (sin x + 2) = 0 or 2 sin x = 1 ⇒ sin x = –2 or sin x = \(\frac{1}{2}\)

Since sin x = – 2 is inadmissible, therefore, sin x = \(\frac{1}{2}\)

⇒ x = 30°, 150°, i.e. x = \(\frac{π}{6}\) ,\(\frac{5π}{6}\).

∴ The number of solutions x ∈[0, π] are 2.