Solve the equation sin2x-sin4x=0?

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Being a subtraction of 2 alike trigonometric functions, we'll transform it into a product.

2 cos [(2x+4x)/2]*sin [(2x-4x)/2]=0

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

We'll put each factor from the product as being 0.

cos 3x=0

This is an elementary equation:

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

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

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

We'll solve the second elementary equation:

sin (-x)=0

-sin (x)=0

sin x=0

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

x=k*pi

The set of solutions:

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

sin2x-sin4x=0

sinx-siny = 2cos(x+y)/2*sin(x-y)/2

==> sin2x-sin4x= 2cos(3x)*sin(-x) = 2cos3x*-sinx= -2cos3xsinx=0

==> -2cos3x*sinx= 0

then cos3x=0 OR sinx=0

cos3x=0

==> 3x= pi/2 + 2n*pi

==> x= pi/6 +(2*pi/3)n (n is integer)

sinx= 0

==> x= n*pi

==> x = {n*pi, pi/6+(2pi/3)n}

To solve sin2x-xsin4x = 0.

Solution:

Sin4x = 2sin2x*cos2x. Therefore the given expression becomes:

sin2x -2sin2x*cos2x = 0. Or

sin2x(1-2 cos2x) = 0.Or

sin2x = 0. Or 1-cos2x = 0.

2x = npi. Or

x = npi/2, n = 0,1,2,....

1-2cos2x = 0 gives cos2x = 1/2. Or 2x = 2npi+ or -pi/3. Or

x = npi+pi/6 or x = npi -pi/6, n = 0,1,2...

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