How do you find the points where the graph of the function #y = (sqrt 3)x + 2 cosx# has horizontal tangents and what is the equation?

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Answer 1 · 2016-03-03T10:34:46

#x = pi/6 , {5pi}/6 , {13pi}/6 , {17pi}/6 ...#

Tangent line Equation

#y = Y#, where #Y# is the #y#-coordinate of the point.

Points with horizontal tangents are point that have #frac{"d"y}{"d"x} = 0#.

#frac{"d"y}{"d"x} = sqrt3 - 2 sinx#

So solve

#sqrt3 - 2 sinx = 0#

#sinx = sqrt3/2#

#x = pi/6 , {5pi}/6 , {13pi}/6 , {17pi}/6 ...#

or equivalently

#x = (2k+1/2)pi +- pi/3#, where #k in ZZ#.

For each value of #x = X# that has a gradient of zero, the horizontal line #y = sqrt3 X + 2cosX# is the tangent line at the respective points.