Question

If `sin2x=0 ⇔ x=(kpi)/2`, why isn't `sin5x=0 ⇔ x=(kpi)/5`?

How is `sin5x=0 ⇔ x=(2kpi)/5 V x=(pi+2kpi)/5`?

Answer 1

`sin(5x) = 0` does imply `x = (k pi)/5` where `k in ZZ`

Explanation:

I am assuming that the V in the expression

`sin5x=0 ⇔ x=(2kpi)/5 V x=(pi+2kpi)/5`

stands for "or", i,e, what was meant was

`sin5x=0 ⇔ x=(2kpi)/5 vv x=(pi+2kpi)/5`

Now `x=(2kpi)/5` means that `x` is an even multiple of `pi/5`, while `x = (pi+2k pi)/5 = ((2k+1)pi)/5` means that `x` is an odd multiple of `pi/5`.

Hence

`x=(2kpi)/5 vv x=(pi+2kpi)/5`

means that `x` is either an even multiple of `pi/5` or an odd multiple of `pi/5` - which simply means that it is a multiple of `pi/5`

Answer 2

There are 3 different solutions in case sin 5x = 0

Explanation:

In fact, in the case (sin 5x = 0), there are 3 different solutions:
a. `5x = 0 + 2kpi` --> `x = (2kpi)/5`
b. `5x = pi + 2kpi` --> `x = (2k + 1)(pi)/5`
c. `5x = 2pi + 2kpi` -->`x = (k + 1)(2pi)/5`

In case k = 0, there are 3 different answers:
a. `x = 0`
b. `x = (pi)/5`
c. `x = (2pi)/5`
In case k = 1, there are 3 answers:
a. `x = (2pi)/5`
b. `x = (3pi)/5`
c. `x = (4pi)/5`