JEE Main 9 Jan 2019 Shift 2 Solved Paper

sin x – sin 2x + sin 3x = 0
⇒ (sin x + sin 3x) – sin 2x = 0
⇒ 2 sin 2x . cos x – sin 2x = 0
⇒ sin 2x (2 cos x – 1) = 0
⇒ sin 2x = 0 or cos x =

1

2

⇒ x = 0 ,

π

3

there are two values of given equation