2sin^2(x) = 2 + cos(x)

interval [0,2pi)

How do I solve this? Help would be greatly appreciated!

1 answer

2(1-cos^2 x) = 2 + cos x

2 - 2 cos^2 x = 2 + cos x
2 cos^2 x + cos x = 0

cos x (2 cos x + 1 ) = 0
cos x = 0 when x = pi/2 or 90 degrees or 3 pi/2 or 270 deg
or
cos x = -1/2 at x = 2pi/3 or 120 degrees or at 4 pi/3 or 240 degrees
Similar Questions
  1. Solve the equation on the interval [0,2pi).2sin^2x-3sinx+1=0 (2sinx+1)(sinx+1) I don't think I did the factoring correctly. When
    1. answers icon 2 answers
  2. Solve the equation in the interval [0, 2pi].2sin(t)cos(t)-cos(t)+2sin(t)-1=0 Answer needs to be formatted as x={Insert answers
    1. answers icon 1 answer
    1. answers icon 1 answer
  3. 2sin(x)cos(x)+cos(x)=0I'm looking for exact value solutions in [0, 3π] So I need to find general solutions to solve the
    1. answers icon 6 answers
more similar questions