Cheyenne B.

asked • 05/28/16

Solve for 2cos-sin^2x=cos^2x over the interval [0,2pi)

Solve for 2cosx - sin2x= cos2x over the interval [0,2pi)
1.) x=pi/6 and 11pi/6
2.) x=5pi/6 and 7pi/6
3.) x=2pi/3 and 4pi/3
4.) x=pi/3 and 5pi/3

1 Expert Answer


Ira S. answered • 05/28/16

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