# Solve the equation sin4x+sin2x=0.

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sin4x+sin2x=0

But we know that, sinx+siny= 2sin(x+y)/2 *cos(x-y)/2

==> sin4x+sin2x = 2sin3x*cosx = 0

2sin3x*cosx=0

==> sin3x = 0 OR cosx=0

sin3x=0

==> 3x= n*pi

==> x= n*pi/3

cosx= 0

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

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

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

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

2 sin 3x * cos x=0

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

sin 3x=0

This is an elementary equation:

3x = (-1)^k*arcsin 0 + k*pi

3x=(-1)^k*0+k*pi

x=k*pi/3, where k is an integer number.

We'll solve the second elementary equation:

cos x=0

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

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

The set of solutions:

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

To solve sin4x+sin2x = 0.

Solution:

WE know that 2sin2A = sinA*cosA. Therefore,

Sin4x = 2sin2x cos2x.

Therefore, the equation,

sin4x+sin2x = 0 becomes 2sin2xcos2x+sin2x = 0.

sin2x(2cos2x +1) = 0.

Or sin2x = 0, gives 2x = npi . Or x = npi/2, where n = 1,2...

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

x = npi +or- pi/3 , n =1,2,3....