Help please, F(x)= sin(3 cos(x) ) ... find exact values of x where the horizontal tangents are. HOW?!?!?

Question

I don't understand how to do this. Thank you!

Answer 1

Horizontal tangents exist at points in which the derivative at the point is 0.

So first find the derivative (using the chain rule):
F'(x) = cos(3cos(x)) * -3sin(x)

Now, set the derivative equal to 0:
cos(3cos(x)) * -3sin(x) = 0

So
(1) cos(3cos(x)) = 0
and
(2) -3sin(x) = 0

(1)
cos(3cos(x)) = 0
3cos(x) = nπ/2, where n is an odd integer
cos(x) = nπ/6
x = arccos(nπ/6)

(2)
-3sin(x) = 0
sin(x) = 0
x = nπ, where n is an integer.

So the solutions are:
x = arccos(nπ/6)
and
x = nπ

Answer 2

Horizontal tangents occur at maxima and minima of functions.
F(x)= sin(3 cos(x) )
F'(x)= - (cos(3 cos(x) ))sin(x) = 0 for maxima and/or minima
sin(x) = 0
x = 0, nπ
cos(3 cos(x) ) = 0
3 cos(x) = (2n + 1)π/2
cos(x) = (2n + 1)π/6 = ± π/6
(all other possible values are > 1 or < - 1)
x = arccos(± π/6)
x ≈ nπ ± 1.019726743695450250751405049780
(I'm at a loss as to exact value of these.)
or
x = 0, nπ
Edit: 0 is redundant, being included in nπ

Answer 3

To find the values that produce horizontal tangents, you must 1. find the derivative
2. set it equal to zero and solve for all values that make the deriv = 0

y = sin(3 cosx)

(dy/dx) = (-3sin x) * cos (3 cos x)

now set to zero
(-3sin x) * cos (3 cos x) = 0

Let u = 3cosx, then (-3sin x)*(cos u) is obviously zero when sin x = 0 or when cos u = 0

for sinx = 0
x = n(pi) where n = {0, 1, 2, 3,...}

for cos u = 0
u = n(pi)/2 and u = 3n(pi)/2 for same n as above

but because we have that u = (3 cos x)

3cosx = n(pi)/2 and 3cosx = 3n(pi)/2

x = arccos[ n(pi)/6] and x = arccos[ n(pi)/3]

so the final answers are:

x = n(pi)
x = arccos[ n(pi)/6]
x = arccos[ n(pi)/3]

where n = {0, 1, 2, 3, ...}