M2 maths question

sinx+sin2x+sin3x=02sin2xcosx + 2sin2x=02sin2x (cosx+1)=0Case1: sin2x=0 2x=0,π,2π,3π,4π x=0,π/2,π,3π/2,2πCase2: cosx=-1 x=πTherefore, x=0,π/2,π,3π/2,2π

Sinx+Sin2x+Sin3x=0
Sin2x+Sinx+Sin3x=0
Sin2x+Sin(2x－x)+Sin(2x+x)=0
Sin2x+2Sin2xCosx=0
Sin2x(1+2Cosx)=0
Sin2x=0 or Cosx=－1/2
0<=x<=2π
0<=2x<=4π
Sin2x=0
2x=0 or π or 2π or 3π or 4π
x=0 or π/2 or π or 3π/2 or 2π………………
Cosx=－1/2
x=2π/3 or 4π/3…………………………….
So
x=0 or π/2 or 2π/3 or π or 4π/3 or 3π/2 or 2π

sin(x) + sin(2x) + sin(3x) = 0sin(2x) + 2sin(2x)cos(x) = 0 sin(2x)(1 + 2cos(x)) = 0 * sin(2x) = 0 or cos(x) = - 1/2 x= 0, 2π/3, π/2, π, 4π/3, 3π/2 or 2π * sin(A) + sin(B) =2sin((A+B)/2)cos((A-B)/2)

2013-06-11 13:00:38 補充：
星期八's answer is WRONG.