# solving a trigonometric equation

• Nov 8th 2009, 02:27 PM
satis
solving a trigonometric equation
this should be fairly easy, but I'm afraid as I'm at a loss as on how to solve this.

sinx = cosx on an interval of 0 <= x < 2pi

I can reason it out... since sin is opposite over hypotenuse and cos is adjacent over hypotenuse, this is only true where adjacent = opposite, making it $\frac{\pi}{4}$ and $\frac{5\pi}{4}$.

But how is the problem solved algebraically?
• Nov 8th 2009, 02:30 PM
e^(i*pi)
Quote:

Originally Posted by satis
this should be fairly easy, but I'm afraid as I'm at a loss as on how to solve this.

sinx = cosx on an interval of 0 <= x < 2pi

I can reason it out... since sin is opposite over hypotenuse and cos is adjacent over hypotenuse, this is only true where adjacent = opposite, making it $\frac{\pi}{4}$ and $\frac{5\pi}{4}$.

But how is the problem solved algebraically?

$\frac{sin(x)}{cos(x)} = tan(x) = 1$

This is undefined for x = $\frac{\pi}{2}$
• Nov 8th 2009, 02:44 PM
satis
Oh, that was simple. Thank you.