# Thread: solving a trigonometric equation

1. ## 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 $\displaystyle \frac{\pi}{4}$ and $\displaystyle \frac{5\pi}{4}$.

But how is the problem solved algebraically?

2. 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 $\displaystyle \frac{\pi}{4}$ and $\displaystyle \frac{5\pi}{4}$.

But how is the problem solved algebraically?
$\displaystyle \frac{sin(x)}{cos(x)} = tan(x) = 1$

This is undefined for x = $\displaystyle \frac{\pi}{2}$

3. Oh, that was simple. Thank you.