Given , sinx + sin2x + sin3x + sin4x =0
(sin4x + sin 2x ) + ( sin3x + sinx ) =0
Using , (sinA +sinB) formula =>
2sin(4x+2x)/2 cos(4x-2x)/2 + 2sin(3x+x)/2 cos (3x -x )/2 =0
2 sin 6x/2 cos 2x/2 + 2sin 4x/2 cos2x/2 =0
2 sin3x cos x + 2 sin2x cosx = 0
2 cosx ( sin3x + sin2x ) = 0
2cos x ( 2 sin (3x+2x)/2 cos (3x-2x )/2 ) =0
4 cosx sin 5x/2 cosx/2 = 0
cosx =0 ; sin 5x/2 = 0 ; cos x/2 = 0
x = (2n+1)π/2 ; 5x/2 = nπ ; x/2 = (2n+1)π/2
x = (2n+1)π/2 ; x = 2nπ/5 ; x = (2n+1) π
