Rolls Theorem (trig functions)

[Jonathan_Kyle]

Homework Statement

f(x) = sin5x ; [π/5,2π/5] finding the point c which f'(x) =0. I understand the theorem and how to complete it, my issue is using the triq functions

Homework Equations

f'(x) = 5cos5x

The Attempt at a Solution

5cos5x=0
cos5x=0
5x=π/3
x=π/15
my answer is not correct, I am using mathlab and I cannot see how they came to their answer

Edit: Mathlab answer is 3π/10

 

[AdityaDev]

cos5x=0
5x=π/2 or 3π/2
x=π/10 or 3π/10
x=π/10 is not in domain so answer is 3π/10

 

[Jonathan_Kyle]

Thank you for the reply, I am obviously doing something wrong between;
cos5x=0
and
5x= π/2 or 3π/2 ( i understand where you get 3π/2, just another revolution from π/2)
what i don't understand is how you got π/2, I was under the impression that you take the inverse of cosine from each side of the equation to isolate x on the left hand side. Furthermore, I was told that finding the inverse of the cosine function is π/3, because that gives you 1/2. I think I was just given some back information on how to solve that portion of the problem.

 

[SteamKing]

Thank you for the reply, I am obviously doing something wrong between;
cos5x=0
and
5x= π/2 or 3π/2 ( i understand where you get 3π/2, just another revolution from π/2)

Not quite. Remember, 1 revolution = 2π radians, so it's a half revolution.

what i don't understand is how you got π/2, I was under the impression that you take the inverse of cosine from each side of the equation to isolate x on the left hand side. Furthermore, I was told that finding the inverse of the cosine function is π/3, because that gives you 1/2. I think I was just given some back information on how to solve that portion of the problem.

If you were trying to find θ such that cos θ = 0, what would be the value of θ in this case? For what angles is the cosine zero?

You can use the unit circle to figure this out, and you should have memorized where sin θ = 0, cos θ = 0, tan θ = 1, etc.