# Solve the equation : sin5x - sinx = 0Solve the equation : sin5x - sinx = 0

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sin5x - sinx = 0

We know that:

sina - sinb = 2cos(a+b)/2*sin(a-b)/2

==> sin5x - sinx = 2*cos(3x)*sin(2x) = 0

==> 2cos3x*sin2x = 0

Divide by 2:

==> cos(3x)*sin(2x) = 0

Then we have two cases:

cos(3x)= 0 OR sin(2x) = 0

when cos(3x) = 0

==> 3x = pi/2 , 3pi/2

==>** x= pi/6 , pi/2 **

When sin2x = 0

==> 2x= pi, 2pi

**==> x= pi/2 , pi.**

**==> x= { pi/6+ 2npi , pi/2+2npi , pi + 2npi}**

sin5x -sinx = 0.

To solve for x.

We know sin5x- sinx = 2sin (5x+x)/2 *cos(5x-x)/2

So we rewrite the equation as:

2 sin3x*co2x = 0

Therefore by zero product rule, sin3x = 0. Or cos2x= 0

sin3x = 0 gives: 3x = npi. Or x = npi/3, n= 0, 1,2...

cos2x = 0 gives 2x = 2npi+or-pi/2 Or

x = (npi +pi/4) or (npi-pi/4), n = 0,1,2,.....

To solve the equation, we'll have to transform the difference of 2 like trigonometric functions into a product.

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

2 cos 3x * sin 2x=0

We'll set each factor from the product as 0.

cos 3x = 0

This is an elementary equation:

3x = +/-arccos 0 + 2*k*pi

3x = +/- pi/2 + 2*k*pi

We'll divide by 3 both sides:

x = +/- pi/6 + 2*k*pi/3, where k is an integer number.

We'll solve the second elementary equation:

sin 2x=0

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

2x = 0 + k*pi

We'll divide by 2 both sides:

x = k*pi/2

The set of solutions:

**S={k*pi/2}or{+/-pi/6 + 2*k*pi/3}**