How do you find the points where the graph of the function #f(x)=2cosx+sin2x# has horizontal tangents and what is the equation?

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Answer 1 · 2016-11-13T01:45:33

The function will have horizontal tangents at #x = pi/6#, #x = (5pi)/6# and #x= (3pi)/2#.

The tangent will be horizontal when the slope (or the derivative) equals #0#. By the product and chain rules, we have:

#f'(x) = -2sinx + 2cos(2x)#

Set this to #0# and solve for #x#.

#0 = -2sinx + 2cos(2x)#

#0 = -2sinx + 2(1- 2sin^2x)#

#0= -2sinx+ 2 - 4sin^2x#

#4sin^2x + 2sinx - 2 = 0#

#2(2sin^2x + sinx - 1) = 0#

Let #t= sinx#.

#2t^2+ t - 1= 0 #

#2t^2 + 2t - t - 1 = 0#

#2t(t + 1) - (t + 1) = 0#

#(2t - 1)(t + 1) = 0#

#t = 1/2 and t = -1#

#sinx = 1/2 and sinx= -1#

#x = pi/6, (5pi)/6 and (3pi)/2#

Note that these points are only in the interval #[0, 2pi]#.

Hopefully this helps!