Question

Solve sin x + sin 2x + sin 3x = cos x + cos 2x + cos 3x.

Answer 1

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

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 .

Answer 2

Solution :

sinx+sin2x+sin3x = cos x + cos2x+cos3x

=> sinx+sin3x+sin2x = cosx+cos3x+cos2x

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

+ 2cos[(x+3x)/2]cos[(x-3x)/2] + cos 2x

=>2sin2xcosx+sin2x=2cos2xcosx+cos2x

=> sin2x( 2cosx + 1 ) = cos2x( 2cosx + 1 )

=> sin2x(2cosx+1) - cos2x(2cosx+1) = 0

=> (2cosx+1)( sin2x - cos2x ) = 0

Therefore,

2cosx+1 = 0 or sin2x = cos 2x

cos x = -1/2 or ( sin2x)/(cos2x) = 1

cos x = (-1/2 ) or tan 2x = 1

x = 2nπ±2π/3 or x = (nπ/2)+(π/8) , n€Z

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