# Find x, anyone help?

• May 10th 2008, 02:28 PM
LaurenB
Find x, anyone help?
sinx - x multiplied by cosx - (pi/2)=0

any method to give the solution thats not an approximation

thanks
• May 10th 2008, 02:48 PM
mr fantastic
Quote:

Originally Posted by LaurenB
sinx - x multiplied by cosx - (pi/2)=0

any method to give the solution thats not an approximation

thanks

Well the method I propose uses routine algebra and recognition of special angles and values.

Using the null factor law (you might know it by a different name):

$\sin (x) - x = 0 \Rightarrow \sin x = x$ .... (1)

or

$\cos (x) - \frac{\pi}{2} = 0 \Rightarrow \cos x = \frac{\pi}{2}$ .... (2)

Regarding equation (1), an obvious exact solution is x = 0. The tricky bit is to show that this actually the only solution ......

Edit: Deleted my remarks on equation (2) for reasons made obvious by an ancient Greek philosopher!
• May 10th 2008, 03:29 PM
Plato
Quote:

Originally Posted by mr fantastic
$\cos (x) - \frac{\pi}{2} = 0 \Rightarrow \cos x = \frac{\pi}{2}$ .... (2)
Equation (2) is obviously your best chance for exact solutions: $x = \pm \frac{\pi}{2}, ~ \pm \frac{3\pi}{2}, ~ \pm \frac{5\pi}{2} ......$ and in general $x = \frac{(2n-1) \pi}{2}$ where n is any integer.

Do you want to rethink that answer?
Maybe $\arccos \left( {\frac{\pi }{2}} \right)$?
• May 10th 2008, 03:37 PM
mr fantastic
Quote:

Originally Posted by Plato
Do you want to rethink that answer?
Maybe $\arccos \left( {\frac{\pi }{2}} \right)$?

(Doh) (Rofl)

Thankyou.

And $\frac{\pi}{2} > 1$ leads to the obvious conclusion .......