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I'm solving this equation:

$$\sin(3x) = 0$$

The angle is equal to 0, therefore:

$$3x=0+2k\pi \space\vee\space3x= (\pi-0)+2k\pi$$

$$x = \frac {2}{3}k\pi \space \vee \space x = \frac {\pi + 2k\pi}{3}$$

Though, the answer is

$$x = k\frac {\pi}{3}$$

It looks like the two trigonometric equations have been combined into one. I must have made a mistake. Any hints?

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    $\begingroup$ Notice that for $\sin x = 0$, the solutions $x = 0 + 2k\pi \; \vee \; x = \pi + 2k\pi$ can be combined as $x = k\pi$ (where $k \in \mathbb{Z}$). Draw the solutions and realise that you're not "missing" anything: both ways of writing down the solutions contain the exact same angles. $\endgroup$
    – StackTD
    Feb 12, 2016 at 13:07
  • $\begingroup$ Thanks a lot @StackTD. This makes a lot of sense. Feel free to post your comment as an answer and I'll be glad to accept it. $\endgroup$
    – Cesare
    Feb 12, 2016 at 13:14
  • $\begingroup$ Thanks; I elaborated a bit in the answer. $\endgroup$
    – StackTD
    Feb 12, 2016 at 13:18
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    $\begingroup$ Notice that:$$\left\{\frac23k\pi:k\in\Bbb Z\right\}=\left\{\dots,-\frac43,-\frac23,\frac23,\frac43,\dots\right\}$$and:$$\{ \frac{\pi+2k\pi}3:k\in\Bbb Z\} =\\\left\{\dots,-\frac33\pi,-\frac13\pi,\frac13\pi, \frac33\pi,\dots\right\}$$ $\endgroup$ Feb 12, 2016 at 13:30

3 Answers 3

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All we need is for $3x$ to be an integer multiple of $\pi$. In other words

$$3x = k\pi \Rightarrow x = \frac{k\pi}{3}$$

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You're not missing anything: sometimes it's possible to efficiently combine sets of solutions.

Notice that for $\sin x=0$, the solutions $x=0+2k\pi$ ($0$ and then 'adding full circles') $\vee \; x=\pi+2k\pi$ ($\pi$ and then 'adding full circles') can be combined as $x=k\pi$ ($0$ and 'adding half circles'), where always $k \in \mathbb{Z}$.

Draw the solutions and realise that you're not 'missing' anything: both ways of writing down the solutions contain the exact same angles; you 'run through' the same angles.

Addendum

This is not always possible for equations of the form $\sin x = c$ (only if $c=k\pi$) or $\cos x = c$ (only if $c=\pi/2+k\pi$), but it is always possible for $\tan x = c$ since the solutions $$x = \arctan c + 2k\pi \, \vee x = \pi + \arctan c + 2k\pi$$can alwayes be combined as $$x = \arctan c + k\pi$$ You can easily see this by drawing a trigonometric circle and visualising the solutions.

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Your solutions are: $$ x = \frac{\pi}{3}(2k),\phantom{NNNNNNNN} x = \frac{\pi}{3}(2k + 1). $$ So, you've shown that $x$ is either $\frac{\pi}{3}$ times an even integer, or else $\frac{\pi}{3}$ times an odd integer. In other words, $x$ is $\frac{\pi}{3}$ times an integer.

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