# What is the sollution of sin2x-sin3x?I got two solutions +-Pi/5 and +-3Pi/5 ...but I can't get one more, that is kPi...

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sin(A+B) = sinAcosB + sinBcosA

sin(A-B) = sinAcosB - sinBcosA

so sin(A+B) - sin(A-B) = 2sinBcosA

In this case A+B = 2x and A-B = 3x

A = 5/2 x and B = -1/2 x

so

sin2x - sin3x = 2cos(5/2x)sin(-1/2x) = -2cos(5/2x)sin(1/2x) = 0

We will now use the zero product property that if a*b=0 then either a=0 or b=0

So either cos(5/2x) = 0 or sin(1/2x) = 0

cos(y) = 0 when y = 1/2pi + pi(n) so cos(5/2x) = 0 when 5/2x = 1/2pi + pi(n) which gives x = 1/5pi + 2/5pi(n)

sin(y) = 0 when y = pi(n) so sin(1/2x) = 0 when 1/2x = pi(n) which gives x = 2pi(n)

**So our solution is x = {(2n+1)pi/5, 2pi(n) | n is an integer}**

We'll re-write what the enunciation provided in a proper manner:

sin 2x - sin 3x = 0

We'll transform the difference of sines into a product, to solve the equation:

sin a - sin b = 2cos[(a+b)/2]*sin[(a-b)/2]

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

sin 2x - sin 3x = 2cos[(5x)/2]*sin[(-x)/2]

We'll re-write the equation:

2cos[(5x)/2]*sin[(-x)/2] = 0

We'll divide by 2:

cos[(5x)/2]*sin[(-x)/2] = 0

We'll cancel each factor:

cos[(5x)/2] = 0

5x/2 = +/-arccos 0 + 2kpi

5x/2 = +/-(pi/2) + 2kpi

5x = +/-(pi) + 4kpi

x = +/-(pi/5) + 4kpi/5

sin[(-x)/2] = 0

x/2 = (-1)^k*arcsin 0 + kpi

x/2 = kpi

x = 2kpi

**The solutions of the equation are: {-(pi/5) + 4kpi/5}U{(pi/5) + 4kpi/5}U{2kpi}.**