# Solve for x : sin4x-sin2x=0

*print*Print*list*Cite

### 2 Answers

For, the beginning, we'll express sin 4x = sin 2*(2x)

But, the formula for the double angle sin 2a = 2sina*cosa.

We'll aply the formula to sin 2*(2x):

sin 2*(2x) = 2 sin 2x* cos 2x (1)

Now, we'll re-write the equation, substituting sin 4x by the formula (1):

sin 4x - sin 2x = 0

2 sin 2x* cos 2x - sin 2x = 0

We'll factorize by sin 2x:

sin 2x*(2cos 2x - 1) = 0

We'll set each factor as 0:

sin 2x = 0

2x = +/-arcsin 0

2x = 0

**x = 0**

We'll set 2cos 2x - 1 as 0:

2cos 2x - 1 = 0

We'll add 1 both sides:

2cos 2x = 1

We'll divide by 2:

cos 2x = 1/2

2x = +/- arccos (1/2)

2x = +/- (pi/3)

We'll divide by 2:

**x = +/- pi/6**

**The solution -pi/6 could be expressed as x = 2pi - pi/6**

**x = 11pi/6**

**The solutions of the equation, over the interval [0,2pi], are:**

**{0 ; pi/6, 11pi/6}.**

To solve for x . sin4x -sin2x = 0.

To solve the equation we use sin2A = 2sinAcosA.

Therefore , sin4x = sin2(2x) = 2sin2x*cos2x.

So we substitute sin4x = 2sin2x cos2x 9n the given equation:

2sin2x*cos2x - sin2x = 0

sin2x*(2cos2x -1) = 0..............(1).

Again we substitute cos2x = 1-(2sinx)^2 and sin2x = 2sinxcosx on (1):

2sinx*cosx {2(1-2(sinx)^2) -1} = 0

2 sinx*cosx{1 - 4(sinx)^2} = 0

Equate each non numeriacal factors to zero:

sinx = 0, cosx = 0 , 1-4(sinx)^2 = 0

So sin x = 0 gives: x= 0 or pi

cosx = 0 gives x = pi/2 or 3pi/2

1-4(sinx)^2 = gives: sinx = 1 or =-1. So x = pi/4, 3pi/4 , 5pi/4 or 7pi/4