Originally Posted by

**Moo** Hello,

I don't understand your writing, though the result is correct. But this implication makes no sense...

Let X=5x. Then X is between 0 and 10 pi.

So you have to solve for sin(X)=0

As you said, X=k*pi, with k=..., -2, -1, 0, 1, 2, ....

Since X is between 0 and 10 pi, therefore k=0, 1, ..., 9, 10 (0 and 2 pi are inclusive).

Solutions are $\displaystyle X=0, \ \pi, \ 2 \pi, \dots , \ 9 \pi, \ 10 \pi$

---> $\displaystyle x=0, \frac{\pi}{5}, \frac{2 \pi}{5}, \dots , \frac{9 \pi}{5}, 10 \pi$

Is it ok ? oO