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

neela | Student

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).

william1941 | Student

The slope of the tangent drawn to a curve f(x) at any point a is given by the value of the first derivative of the slope at the point x=a.

Here we need the tangent to be horizontal, so the slope should be equal to 0.

Therefore, we differentiate y=sin(x)+sqrt(3)*cos(x)

dy/ dx = cos x - sqrt 3 * sin x

Equating this to 0

=> cos x - sqrt 3 * sin x = 0

=> cos x = sqrt 3 * sin x

=> tan x = 1/ sqrt 3

So x = arc tan (1/ sqrt 3) = pi / 6 + n*pi

The required values of are pi / 6 + n*pi.