At what points will #y = cos(2x) + sinx# have horizontal tangents?

1 Answer
Oct 25, 2016

#x = arcsin(1/4), pi/2, pi- arcsin(1/4), (3pi)/2#

Explanation:

Let's start by differentiating the function. Before doing this to the complete function, we need to find the derivative of the most complex part, #cos2x#.

Let #y = cosu# and #u = 2x#. #y' = -sinu# and #u' = 2#.

#dy/dx = 2 xx -sinu = -2sin(2x)#

We can now differentiate the entire function.

#y' = -2sin(2x) + (sinx)' = -2sin2x + cosx#

Now, we are looking for horizontal tangents. Horizontal tangents are horizontal because their slope is #0#, or the parameter #m# in #y = mx + b# equals #0#.

The slope of the tangent to a function at a given point #x = a# is given by evaluating #f(a)# into the derivative. We know the slope, so we can solve for #x# instead of solving for #y#.

#0 = -2sin2x + cosx#

Recall that #sin2x = 2sinxcosx#.

#0 = -2(2sinxcosx) + cosx#

#0 = -4sinxcosx + cosx#

#0 = cosx(-4sinx + 1)#

#cosx = 0 and sinx = 1/4#

#x = pi/2, (3pi)/2, pi - arcsin(1/4) and arcsin(1/4)#

Hopefully this helps!