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

Question

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

Answer 1

you can use the pythagorean substitution. Since Cos² x + Sin² x = 1, Sin² x = 1-cos² x.

 

So you have 2 cos x -(1-cos² x) = cos² x

2cos x - 1 + cos² x = cos²x

2cos x -1               = 0

cos x                     = 1/2

 

Since cosine is positive, you have an answer in the first and the other in the fourth quadrant.

You can use a calculator to find the one in the first quadrant, or use the 30-60-90 to find that x = 60 = Π/3.

A 60 degree reference angle in the fourth quadrant would be a 300 degree angle....or 5π/3.

 

Hope this helped.