how do i find the points on y = (cos x)/(2 + sin x) at which the tangent line is horizontal?

Question

um.. this is what i did...

0 = (cos x)/(2 + sin x)

(2 + sin x) (0) = cos x

0 = cos x

x = pi/2 and 3pi/2....

um.. i dont know if i did that right..

thanks!

Answer 1

y = cos x / ( 2 + sin x) . . . . use derivative of fraction
y ' = [ cos x ( cosx) - ( 2 + sin x)(- sin x ) ] / ( 2 + sin x)^2 = 0
cos^2 x + 2 sin x + sin^2 x = 0
2 sin x = -1 . . . . . . . . . . . . since . . . sin^2 x + cos^2 = 1
sin x = - 1/2
x = - 30 or 330 deg . . . . and . . . x = 210 deg

Answer 2

y = (cos x)/(2 + sin x)
y' = [(2+sinx)(-sinx) -(cosx)(cosx)]/(2+sinx)^2
y' = [-2sinx -sin^2x-cos^2x}/(2+sinx)^2
y' = (-2sinx -1)/(2+sinx)^2
Set y' =0 and solve for x getting
sinx = -1/2 --> x = -.5235987756 radians = 330 degrees or
210 degrees = 5pi/3 or 11pi/6 radians.
So tangent line is horizontal at 11pi/6 and 5pi/3 and all angles coterminal with these angles.