# 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?

Jun 3, 2018

$\sin \left(5 x\right) = 0$ does imply $x = \frac{k \pi}{5}$ where $k \in \mathbb{Z}$

#### 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 $x = \frac{2 k \pi}{5}$ means that $x$ is an even multiple of $\frac{\pi}{5}$, while $x = \frac{\pi + 2 k \pi}{5} = \frac{\left(2 k + 1\right) \pi}{5}$ means that $x$ is an odd multiple of $\frac{\pi}{5}$.

Hence

$x = \frac{2 k \pi}{5} \vee x = \frac{\pi + 2 k \pi}{5}$

means that $x$ is either an even multiple of $\frac{\pi}{5}$ or an odd multiple of $\frac{\pi}{5}$ - which simply means that it is a multiple of $\frac{\pi}{5}$

Jun 3, 2018

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. $5 x = 0 + 2 k \pi$ --> $x = \frac{2 k \pi}{5}$
b. $5 x = \pi + 2 k \pi$ --> $x = \left(2 k + 1\right) \frac{\pi}{5}$
c. $5 x = 2 \pi + 2 k \pi$ -->$x = \left(k + 1\right) \frac{2 \pi}{5}$

In case k = 0, there are 3 different answers:
a. $x = 0$
b. $x = \frac{\pi}{5}$
c. $x = \frac{2 \pi}{5}$
In case k = 1, there are 3 answers:
a. $x = \frac{2 \pi}{5}$
b. $x = \frac{3 \pi}{5}$
c. $x = \frac{4 \pi}{5}$