Determine the numbers x between 0 and 2pi where the line tangent to the curve is horizontal. y=sin(x)+sqrt(3)*cos(x)

y = sinx +sqrt3 cosx.

The
tangent at (x1, y1) to the curve y = sinx+(sqrt3) cosx is given
by:

y - y1 =
(dy/dx)(x-x1).

dy/dx =
d/dx{sin(x)+sqrt(3)*cos(x)}.

dy/dx =
cosx+(sqrt3)(-sinx).

When the curve is horizontal, (or
parallel to x axis), dy/dx = 0.

So  cosx - (sqrt3)sinx =
0.

cosx =
(sqrt3)sinx.

cosx/sinx =
sqrt3.

Taking the reciprocals of both sides, we
get:

tanx = 1/sqrt3.

x = 30 de
or x = (180+30) deg = 210 degree.

x = pi/6 radians, ot x =
pi+pi/6 = 7pi/6.

Therefore  x = pi/6 or x = 7pi/6 are the
solutions in (0 ,2pi).