How do you find all points on the graph of the function #f(x)=2sinx+sin^2x# at which the tangent line is horizontal?

1 Answer
Nov 19, 2016

The points are #x = pi/2 + 2pin and (3pi)/2 + 2pin#.

Explanation:

We know that the derivative represents the instantaneous rate of change of a function, or the slope of the function. We also know that the slope of a line that is horizontal is #0#. We start by differentiating the function.

#f(x) = 2sinx + sin^2x#

#f(x) = 2sinx + (sinx)(sinx)#

#f'(x) = (0 xx sinx + 2 xx cosx) + (cosxsinx + cosxsinx)#

#f'(x) = 2cosx + 2cosxsinx#

We set this to #0# and solve for #x#.

#0 = 2cosx + 2cosxsinx#

#0 = 2(cosx + cosxsinx)#

#0 = cosx(1 + sinx)#

#cosx = 0 and sinx= -1#

#x = pi/2 + 2pin, (3pi)/2 + 2pin#

Hopefully this helps!