1. ## Inequality

Can somebody show me how to finish this one off? I know that the answer is $\displaystyle [0, \frac{\pi}{4})$ and $\displaystyle (\frac{5\pi}{4},2\pi]$, but I don't know how to lay this out. Here is what I have:

$\displaystyle \sin{x} > \cos{x}$
$\displaystyle \sin{x} - \sin{(\frac{\pi}{2}-x}) > 0$

Not sure how to solve this algebraically.

Thanks.

2. Think about solving this $\displaystyle \sin{x}=\cos{x}$

$\displaystyle x=\frac{\pi}{4}+\pi k \ \ \ k\in\mathbb{Z}$

Now, if we restrict the solutions to the unit circle, we have $\displaystyle x=\frac{\pi}{4} \ \ \frac{5\pi}{4}$

Now, test point to the left and right of those numbers.

3. Originally Posted by dwsmith
Think about solving this $\displaystyle \sin{x}=\cos{x}$
This is a good thought. It may be easier to see where dwsmith found he's solution from by the following

$\displaystyle \sin{x}=\cos{x}$

$\displaystyle \frac{\sin{x}}{\cos{x}}=1$

$\displaystyle \tan{x}=1\implies x=\frac{\pi}{4}+\pi k \ \ \ k\in\mathbb{Z}$