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Find the solution of the equation sin^2 2x - 2sin 4x+3cos^2 2x = 0

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lillyan | Student, Undergraduate | (Level 2) Honors

Posted July 6, 2011 at 11:15 PM via web

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Find the solution of the equation sin^2 2x - 2sin 4x+3cos^2 2x = 0

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted July 6, 2011 at 11:20 PM (Answer #1)

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First, we'll use the double angle identity to re-write the middle term.

sin 4x = sin 2(2x) = 2sin 2x*cos 2x

We'll re-write the equation:

(sin 2x)^2 - 4 sin 2x*cos 2x + 3(cos 2x)^2 = 0

We'll divide by (cos 2x)^2 the equation:

(tan 2x)^2 - 4 tan 2x + 3 = 0

We'll substitute tan 2x by t.

We'll re-write the equation:

t^2 - 4t + 3 = 0

We'll apply quadratic formula:

t1 = [4+sqrt(16 - 12)]/2

t1 = (4+2)/2

t1 = 3

t2 = 1

We'll put tan 2x = t1 => tan 2x = 1 => 2x = arctan1 + k*pi

2x = pi/4 + k*pi

We'll divide by 2:

x = pi/8 + k*pi/2

Let tan 2x = t2 => tan 2x = 3 => 2x = arctan3 + k*pi

x = [arctan(3)]/2 + k*pi/2

The requested solutions of the equation are: {pi/8 + k*pi/2 / k integer number}U{[arctan(3)]/2 + k*pi/2 / k integer number}.

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