# Find all points on graph of f(x)=2sinx+sin^2x where slope = 0

1. Jul 27, 2008

### illjazz

1. The problem statement, all variables and given/known data
Find all points on the graph of the function

$$f(x)=2sinx+sin^2x$$

at which the tangent line is horizontal.

2. Relevant equations
- Power rule
- Chain rule
- Product rule?

3. The attempt at a solution
So I want all points at which the tangent line to this function has a slope of zero.

$$f(x)=2sinx+sin^2x$$

$$f'(x)=2cosx+(sinx)^2$$

$$f'(x)=2cosx+2sinx*\frac{d}{dx}sinx$$

$$f'(x)=2cosx+2sinx*cosx$$

$$f'(x)=cosx(2+2sinx)$$

...?

How do I go from here? I know to set that last equation equal to zero and find all the solutions, but just how do I find those solutions?

The solution from the book says:
$$(\frac{\pi}{2}+2n\pi,3),(\frac{3\pi}{2}+2n\pi,-1)$$ where n an integer

2. Jul 27, 2008

### rock.freak667

Slope=0 means f'(x)=0

then you use the general rule that if AB=0 then,A=0,B=0

3. Jul 27, 2008

### illjazz

Right, I know.. but I don't see how to arrive at those values given in the solution. Is my calculator required here?

4. Jul 27, 2008

First, think back to algebra when you had those equations like (x-4)(x-2)=0, and you solved them by setting up two possibilities: either x-4=0, or x-2=0. You can use the same principle here.

If $$cosx=0$$, or if $$2+2sinx=0$$, then $$f'(x)=0$$.

Now simply find all the points where this can happen. You don't need a calculator - simple trigonometry can help. Where is $$cosx=0$$? And where is $$sinx=-1$$?

5. Jul 27, 2008

### tiny-tim

Hi illjazz!

(have a pi: π and a squared: ² )

I think you're misunderstanding the notation in the solution.

sin(π/2 + 2nπ) = sin(π/2) = 1, and so f(π/2 + 2nπ) = 3.

sin(3π/2 + 2nπ) = sin(3π/2) = … ?, and so f(3π/2 + 2nπ) = … ?.

6. Jul 27, 2008

### HallsofIvy

Staff Emeritus
No, it is not. I know what you mean but this is terrible notation. You mean
$$f'(x)= 2 cosx+ \frac{d}{dx} (sin x)^2$$

Okay, now that is correct.

But why did you do this? Just solve 2cos x+ 2 sin x cos x= 0. That is the same as
2 cos x (1+ 2 sin x)= 0 which means either cos x= 0 or 1+ 2 sin x= 0. That is easy to solve.