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Solve the equation sin 4x - sin 2x = 0

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andrzej | Student, College Freshman | (Level 1) Honors

Posted May 31, 2010 at 11:44 PM via web

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Solve the equation sin 4x - sin 2x = 0

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted May 31, 2010 at 11:49 PM (Answer #1)

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As we can see, we have to calculate a subtraction of 2 alike trigonometric functions, which we'll transform it into a product.

2 cos [(4x+2x)/2]*sin [(4x-2x)/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

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

x = k*pi

The set of solutions:

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

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neela | High School Teacher | (Level 3) Valedictorian

Posted May 31, 2010 at 11:51 PM (Answer #2)

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To solve the equation sin4x-sin2x = 0.

Solution:

sin4x = 2sin2x*cos2x. So the given equation coube be rewriten as:

2sin2x*cos2x-sin2x = 0. Or

sin2x(2cos2x - 1) = 0. Or

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

The other factor 2cos2x -1 = 0 gives: cos2x = 1/2 or

2x = 2npi + (-1)^n* (pi/3). Or

x = npi +(-1)pi/6

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