# Thread: Find x, anyone help?

1. ## Find x, anyone help?

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

any method to give the solution thats not an approximation

thanks

2. 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!

3. 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 $\arccos \left( {\frac{\pi }{2}} \right)$?

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

Thankyou.

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