What are the solutions to the trigonometric equation on the interval [0,2π)?
Select all correct solutions.
2cos^2x=cosx Divide both sides of th equation by cos
2 cos x = 1
cos x = 1/2
x = pi/3 5 pi / 3
Need to be careful about dividing out possible solutions.........
2cos^2x = cos x
2cos^2 x - cos x = 0
cos x ( 2cos x - 1) = 0
Set each factor to 0 , and solve for x
cos x = 0 x = pi/2 and x = 3pi/2 2cos x - 1 = 0
2cosx = 1
cosx = 1/2 x = pi/3 and x = 5pi/3