# Thread: Turning points of y = a sin(2x) + 2 cos(x)

1. ## Turning points of y = a sin(2x) + 2 cos(x)

1)
Given function: $\displaystyle y=a\sin(2x)+2\cos(x)$ between $\displaystyle 0\leq x \leq \pi$
The function has a stationary point on $\displaystyle x=\frac{5}{6}\pi$

A) Calculate a.
B) Find the vertex of the function in this domain.

2. Originally Posted by mathwhat
1)
Given function: $\displaystyle y=a\sin(2x)+2\cos(x)$ between $\displaystyle 0\leq x \leq \pi$
The function has a stationary point on $\displaystyle x=\frac{5}{6}\pi$

A) Calculate a.
B) Find the vertex of the function in this domain.
Start by finding dy/dx. Can you do that?

3. Originally Posted by mathwhat
1)
Given function: $\displaystyle y=a\sin(2x)+2\cos(x)$ between $\displaystyle 0\leq x \leq \pi$
The function has a stationary point on $\displaystyle x=\frac{5}{6}\pi$

A) Calculate a.
B) Find the vertex of the function in this domain.
$\displaystyle y' \left(\frac{5\pi}{6}\right) = 0$

Use the chain rule to differentiate y.

4. i already found the a, a=1
my problem is with B.
i got:
$\displaystyle y'=2\cos(2x)-2\sin(x)=0$
but i dont getting $\displaystyle \frac{5}{6}\pi$ as answer.

5. Originally Posted by mathwhat
i already found the a, a=1
my problem is with B.
i got:
$\displaystyle y'=2\cos(2x)-2\sin(x)=0$
but i dont getting $\displaystyle \frac{5}{6}\pi$ as answer.
It saves time and effort if you say in your question what you can already do.

For B) your job is to find the value of y when x = pi/6. Where are you stuck?

6. in finding the x from: $\displaystyle y'=2\cos(2x)-2\sin(x)=0$

7. Originally Posted by mathwhat
in finding the x from: $\displaystyle y'=2\cos(2x)-2\sin(x)=0$
You asked this question here: http://www.mathhelpforum.com/math-he...tml#post517004

and got a reply. The reply you were given had a simple error in it that you should have realised if you followed the working closely.