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Solve 2 cos3x cos2x = 1 - 2sin3xsin2x on the interval [0, 2pi)

Calculus
Tutor: None Selected Time limit: 0 Hours

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Jan 28th, 2015

Use the formula cos(A - B) = cosAcosB + sinAsinB

2 cos3x cos2x = 1 - 2sin3xsin2x 


2 ( sin 3xsin2x + cos3xcos2x) =1
cos(3x-2x) =1/2
cos x = 0.5

x = cos^-1 ( 0.5) = 60 and 300 degrees

so, x =  pi/3 and 5pi/3

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Jan 28th, 2015

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Jan 28th, 2015
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Jan 28th, 2015
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