Hello,

I don't understand your writing, though the result is (almost) correct. But the "therefore" part makes no sense...sin5x=0 implies 5x=kpie , k=0 to 9

implies x=kpie/5 therefore k=0 to9

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

--->

Is it ok ? oO