Solve sin2x-sin4x+sin6x.?

sin(2x) - sin(4x) + sin(6x)
= sin(4x - 2x) + sin(4x + 2x) - sin(4x)
= sin(4x)cos(2x) - cos(4x)sin(2x) + sin(4x)cos(2x) + cos(4x)sin(2x) - sin(4x)
= 2sin(4x)cos(2x) - sin(4x)
= sin(4x)(2cos(2x) - 1)

If this is an equation equalling 0, then on the interval 0 <= x < 2pi:

sin(4x) = 0 or cos(2x) = 1/2
4x = 0, pi, 2pi, 3pi, 4pi, 5pi, 6pi, 7pi, or 2x = pi/3, 2pi/3, 7pi/3, or 8pi/3
x = 0, pi/4, pi/2, 3pi/4, pi, 5pi/4, 3pi/2, 7pi/4, pi/6, pi/3, 7pi/6, or 4pi/3

(sin6x +sin2x )-sin4x
using [sina + sinb = 2sin{(a+b)/2}cos{(a-b)/2}]
2sin4xcos2x -sin4x
sin4x{ 2cos2x -1}
using {cos2x = 1 - 2sin^(2)x}
sin4x{2 -4sin^(2)x -1}
sin4x{1- 4sin^(2)x}

Do you know this identity:

sin(a - b) = sin(a).cos(b) - cos(a).sin(b) → adapt it to your case

sin(4x - 2x) = sin(4x).cos(2x) - cos(4x).sin(2x)

sin(2x) = sin(4x).cos(2x) - cos(4x).sin(2x)


Do you know this identity:

sin(a + b) = sin(a).cos(b) + cos(a).sin(b) → adapt it to your case

sin(4x + 2x) = sin(4x).cos(2x) + cos(4x).sin(2x)

sin(6x) = sin(4x).cos(2x) + cos(4x).sin(2x)


Do you know this identity:

cos(a + b) = cos(a).cos(b) - sin(a).sin(b) → adapt it to your case

cos(2x + 2x) = cos(2x).cos(2x) - sin(2x).sin(2x)

cos(4x) = cos²(2x) - sin²(2x)

cos(4x) = [1 - sin²(2x)] - sin²(2x)

cos(4x) = 1 - sin²(2x) - sin²(2x)

cos(4x) = 1 - 2sin²(2x) ← memorize this result



= sin(2x) - sin(4x) + sin(6x)

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

= sin(2x) - sin(4x).cos(2x) + cos(4x).sin(2x) + sin(4x).cos(2x) + cos(4x).sin(2x)

= sin(2x) + 2cos(4x).sin(2x) → recall: cos(4x) = 1 - 2sin²(2x)

= sin(2x) + 2[1 - 2sin²(2x)].sin(2x)

= sin(2x) + 2sin(2x) - 4sin³(2x)

= 3sin(2x) - 4sin³(2x)



If you want to solve the equation: sin(2x) - sin(4x) + sin(6x) = 0, it's similar to solve :

3sin(2x) - 4sin³(2x) = 0

sin(2x).[3 - 4sin²(2x)] = 0


First possibility:

sin(2x) = 0

x = 0


Second possibility:

3 - 4sin²(2x) = 0

sin²(2x) = 3/4

sin(2x) = ± (√3)/2


First case: sin(2x) = (√3)/2

2x = π/3 → x = π/6

2x = π - (π/3) → 2x = 2π/3 → x = π/3


Second case: sin(2x) = - (√3)/2

2x = π + (π/3) → 2x = 4π/3 → x = 2π/3

2x = 2π - (π/3) → 2x = 5π/3 → x = 5π/6


→ Solution = { 0 ; π/6 ; π/3 ; 2π/3 : 5π/6 }

2sin4xsin2x-sin4x=sin4x(2sin2x-1)=2sin2xcos2x(4sin^2x-2)

I know this without knowing it is 24