Question

Critical numbers for cos9x=0 and sin9x=0?

Answer 1

Do you mean find the critical numbers for f(x)=cos(9x) and g(x)=sin(9x)?

Answer 2

Yep! @myininaya

Answer 3

You need to start by differentiating f and g.

Answer 4

Well the original equation was cos(9x)^2 and then I used the chain rule to get -18cos(9x)sin(9x) and now I'm trying to find the critical numbers. @myininaya

Answer 5

So we have something like find the critical numbers for \[p(x)=(\cos(9x))^2\]
Your p' is correct.
\[p'(x)=-18\cos(9x)\sin(9x)\]
Now set p'=0
Since we have a*b*c=0
this means a=0 or b=0 or c=0
-18=0 or cos(9x)=0 or sin(9x)=0
But -18 is never equal to 0
But we can solve the other two equations for x since cos and sin do be 0 sometimes
actually infinitely many times do they be 0
So first cos(u)=0
For what values of u is cos 0?

Answer 6

You might find the unit circle handy to look at.

Answer 7

What is cos(pi/2)?
cos(pi/2+pi)?
cos(3pi/2+pi)?
I hope you see a pattern here.

Answer 8

Here I will show you how to solve this first equation but I want you to try to solve the sin(9x)=0.

cos(9x)=0
cos(u)=0 when u=pi/2+n*pi where n is an integer
But in place of 9x we put u
so since u=9x then x=u/9
So if we want an answer in terms of x we say u=x/9=pi/2+n*pi
And solve the equation for x by multiplying 9 on both sides
like so
x/9=pi/2+n*pi
x=9pi/2+9n*pi

Answer 9

You try solving sin(9x)=0

Answer 10

lol googled this myself seen this question and seen a mistake :p
and noticed a mistake
I should have replaced u with 9x
since u is 9x
9x=pi/2+npi
divide both sides by 9
x=pi/18+npi/9

:p

Answer 11

but also we didn't even need the derivative