Have to find the stationary points on the graph y = 3sin^2x

[Zander Forrester]

Homework Statement

Homework Equations

The Attempt at a Solution

[/B]
y = 3sin^2x
dy/dx = 3(2sinxcosx)
dy/dx = 6sinxcosx
For stationary points dy/dx = 0
6sinxcosx = 0
Can you help me from this point please?

 

[Orodruin]

You have three factors, 6, sin(x) and cos(x). What must be true for their product to be equal to zero?

 

[Zander Forrester]

Hello Orodruin,
Thank you for your help.
sin(x) = 0 x = Pi
cos(x) = 0 x = pi/2 and 3pi/2

 

[Orodruin]

sin(x) = 0 x = Pi

This is one of many solutions.

cos(x) = 0 x = pi/2 and 3pi/2

These are two of many solutions.

Unless you specify the allowed region of x, you need to account for all solutions.

 

[Zander Forrester]

Hello again Orodruin,
Region was from 0 to 2pi.
Thank you again for your help.

 

[Mark44]

Homework Statement

Homework Equations

The Attempt at a Solution

[/B]
y = 3sin^2x
dy/dx = 3(2sinxcosx)
dy/dx = 6sinxcosx
For stationary points dy/dx = 0
6sinxcosx = 0
Can you help me from this point please?

Your 2nd equation can be written as dy/dx = 3(sin(2x)), so it's just a matter of finding where sin(2x) = 0 on the interval [0, 2π].

 

[Ray Vickson]

Hello again Orodruin,
Region was from 0 to 2pi.
Thank you again for your help.

So, there are a total of 5 solutions; you have given only three.

 

[Zander Forrester]

Thank you again.

 

[epenguin]

Have you ever seen sinxcosx in a formula before?

Maybe you could use it for a neater solution with fewer oversights

 

[Mark44]

So, there are a total of 5 solutions; you have given only three.

Thank you again.

Were you able to find the other two solutions?

 

[Zander Forrester]

Thanks for your reply.
The textbook gave the three answers I submitted.
With sin(2x) = 0
2x = 0, 180, 360, 540 and 720
X = 0,90,180, 270 and 360
X = 0, pi/2, pi, 3pi/2 and 2pi.
Zander

 

[Orodruin]

X = 0, pi/2, pi, 3pi/2 and 2pi.

These look like five solutions to me.