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

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

How is

##### 2 Answers

#### Explanation:

I am assuming that the V in the expression

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

Now

Hence

means that

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.

b.

c.

In case k = 0, there are 3 different answers:

a.

b.

c.

In case k = 1, there are 3 answers:

a.

b.

c.