Calculus (Urgent Help)
posted by Ren .
The graph of y=cos x * ln cos^2x has seven horizontal tangent lines on the interval [0,2pi]. Find the xcoordinate of all points at which these tangent lines occur.

I believe that it is asking you to take the derivative of the given function, then set it equal to zero. Remember that when you take the derivative to use the product rule. Then find all points on the unit circle between 0 and 2pi and you should come out with 7 x values. Hope this helps.

first of all rewrite it as
y = cosx ( 2 ln(cosx) ) using log rules
= 2 cosx ln(cosx)
now use the product rule to find
dy/dx = 2cosx (sinx/cosx) + (2sinx)(ln(cosx))
=  2sinx(1 + ln(cosx) )
= 0 for horizontal tangents
2sinx = 0 or 1 + ln(cosx) = 0
for 2sinx = 0
sinx = 0
x = 0, π , 2π
for 1 + ln(cox) = 0
ln(cosx) = 1
cosx = e^1 = 1/e
x = 1.194 or 2π  1.194
I only found 5 values, don't know how they got the 7 values.
I was going to replace cos^2 x with (cos 2x + 1)/2
but ran into an awful mess trying to solve the derivative for zero
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