Question

The number of values of $x$ between $0$ and $2\pi$ that satisfies the equation $\sin x+\sin 2x+\sin 3x=\cos x+\cos 2x+\cos 3x$ must be-

Answer 1

Given that

sin x + sin 2x + sin 3x = cos x + cos 2x + cos 3x

(sin 3x + sin x) + sin 2x = ( cos 3x + cos x ) + cos 2x

2[sin(3x+x)/2].[cos(3x-x)/2] + sin 2x = 2[cos(3x+x)/2].[cos(3x-x)/2] + cos 2x

2[sin(4x)/2].[cos(2x)/2] + sin 2x = 2[cos(4x)/2].[cos(2x)/2] + cos 2x

2 sin 2x . cos x + sin 2x = 2 cos 2x . cos x + cos 2x

2 sin 2x . cos x + sin 2x - 2 cos 2x . cos x - cos 2x = 0

sin 2x ( 2 cos x + 1) - cos 2x (2 cos x + 1) = 0

( 2 cos x + 1) ( sin 2x - cos 2x ) = 0

⇒

2 cos x + 1 = 0 And sin 2x - cos 2x = 0

2 cos x = - 1 And sin 2x = cos 2x

cos x = - 1/2 And sin 2x/cos 2x = cos 2x/cos 2x

cos x = - 1/2 And tan 2x = 1

For cos x = - 1/2 :-

The Reference Angle is π/3

Cos is -ve in Quadrant II & III

For Quad II :- x = π - π/3 = 2π/3

For Quad III :- x = π + π/3 = 4π/3

For tan 2x = 1 :-

The Reference Angle is π/4

tan is +ve in Quadrant I & III

For Quad I :- 2x = π/4 ⇒ x = π/8

For Quad III :- 2x = π + π/4 = 5π/4 ⇒ x = 5π/8

The Answer is 2π/3 , 4π/3 , π/8 and 5π/8 .