1. Feb 8, 2010

### pip_beard

1. The problem statement, all variables and given/known data
gradient of the stationary points of y=1-2sinx domain 0<x<2pi

2. Relevant equations

3. The attempt at a solution
dy/dx = -2cosx
-2cosx=0......?

Last edited: Feb 8, 2010
2. Feb 8, 2010

### rock.freak667

Gradient of the stationary point? Do you need to find that or the gradient? Because when you put dy/dx=0 it sort of implies what the gradient will be.

3. Feb 8, 2010

### pip_beard

Sorry. the qu reads 'the curve y=1-2sinx has domain 0<x<pi. find the gradients of the curve at the points where the curve crosses the x-axis

4. Feb 8, 2010

### rock.freak667

Ah that is different.

Well you know the gradient function is dy/dx.

You need to find when the curve crosses the x-axis, or when y=0.

5. Feb 8, 2010

### pip_beard

so do i differentiate it and make it = 0?

so i get -2cosx=0

6. Feb 8, 2010

### pip_beard

how to i rearrange this to work it out?

7. Feb 8, 2010

### rock.freak667

No if y=1-2sinx, the curve crosses, the x-axis when y=0

so solve 1-2sinx=0 for x. Can you do that?

8. Feb 8, 2010

-2sinx=1
sinx=-1/2

9. Feb 8, 2010

### rock.freak667

which gives x as ? ( in the given domain that is)

10. Feb 8, 2010

### pip_beard

principle value of -0.523 (but out of range) the pi-(-0.523) = 3.6651

therfore value = 3.6651 and 6.2832?????

11. Feb 8, 2010

### rock.freak667

this should be sinx=1/2.

But how did you get that value for x?

12. Feb 8, 2010

### pip_beard

ohhh... did sin^-1... so ....

x principle = 0.479425
so pi - 0.4879425 = 2.66216

13. Feb 8, 2010

### rock.freak667

you should get x= π/6 and another one, best to use the exact values, I don't have a calculator with me to divide out pi.

14. Feb 8, 2010

### pip_beard

ok. i'll have a play and post back if i cant do it.. Thanks for the inspiration!