What are the solutions to the trigonometric equation on the interval [0,2π)? 2cos^2x=√3 cosx
2cos^2 (x) = √3 cos x divide both sides by 2
cos^2 (x) = (√3/2) cos x
cos^2 (x) - (√3/2) cos (x) = 0 factor as
cos (x) ( cos (x) - (√3/2) ) = 0
Set each factor to 0 and solve
cos x = 0 and this happens at x = 0
And
cos x - √3/2 = 0
cos x = √3/2
And this happens at x = pi/6 and 11pi /6
So the solutions are x = 0 , x = pi/6 and x = 11pi/6