Question

Find the General Solution of 2Sin²x + 3Cosx=0

Class — 11
Subject — Mathematics
Chapter — Trigonometric Functions

Answer 1

2sin²x + 3cosx = 0

⇒ 2(1 - cos²x) + 3cosx = 0

⇒ 2 - 2cos²x + 3cosx = 0

⇒ 2cos²x - 3cosx - 2 = 0

⇒ 2cos²x - 4cosx + cosx - 2 = 0

⇒ 2cosx(cosx - 2) + (cosx - 2) = 0

⇒ (2cosx + 1)(cosx - 2) = 0

HLO..

IPPEM KANARILELO..

⇒ cosx - 1/2, cosx = 2[Not possible]

⇒ cosx = -1/2

⇒ x = 120°,240°

⇒ x = (2π/3) + 2πn, x = (4π/3) + 2πn

Answer 2

Step-by-step explanation:

2(1- Cos²x) + 3Cosx = 0

2 - 2cos²x + 3 cosx = 0

2Cos²x - 3Cosx - 2 = 0

2Cos²x -4 Cosx + Cosx -2 = 0

2Cosx ( Cosx - 2) + 1 ( Cosx - 2) =0

(2Cosx +1) (Cosx -2) = 0

Now,

2Cosx = -1

2Cosx = (2n + 1)π

Cosx = (n+ 1/2)π/2 ✔️

Cosx = 2 (Not possible)

Range [-1,1]