Inequality

**reastland** · January 13th 2011, 03:20 PM

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

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

Not sure how to solve this algebraically.

Thanks.

**dwsmith** · January 13th 2011, 03:27 PM

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

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

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

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

**pickslides** · January 13th 2011, 03:37 PM

Originally Posted by dwsmith

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

This is a good thought. It may be easier to see where dwsmith found he's solution from by the following

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

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

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

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