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Trigonometry Question

Mathematics
Tutor: None Selected Time limit: 1 Day

Find the values of x between 0 degrees and 360 degrees.

              (3cosx-2)(2cosx+1)=0

Nov 29th, 2014

Set each of them equal to zero:

3cosx - 2 = 0

3cosx = 2

cosx = 2/3

x = cos^-1(2/3) = 48.2 degrees.  Cosine will have the same positive value in quadrants I and IV, so x = 360-48.2 =  311.8 is also an answer.


Do the same for the other half:

2cosx + 1 = 0

cosx = -0.5

x = cos^-1(-0.5) = 120 degrees.  Cosine will have the same negative value in quadrants II and III, so x = 180+60 = 240 is also an answer.

x = 48.2 degrees, 120 degrees, 240 degrees, and 311.8 degrees

Nov 29th, 2014

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Nov 29th, 2014
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Nov 29th, 2014
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