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Solve the given trigonometric equation: $$\sin(\omega t)=- \frac 1 2$$

Here is my attempt:

$$\sin(\omega t)=- \frac 1 2 = \sin \biggr (\pi +\dfrac{\pi}{6}\biggr )$$

Which yields

$$\omega t = \dfrac{7\pi }{6}$$

or

$$\sin(\omega t)=- \frac 1 2 = \sin \biggr (2\pi -\dfrac{\pi}{6}\biggr )$$

$$\omega t = \dfrac{11\pi}{6}$$

Is my assumption correct?

Regards!

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  • $\begingroup$ I've fixed a small error. $\endgroup$ – Mr. Maxwell Jul 7 '18 at 21:17
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You also have $$\omega t = \dfrac{-\pi }{6}+2k\pi$$

Thus the solution is $$\omega t = \{\dfrac{-\pi }{6}+2k_1\pi:k_1 \in Z\}\cup \{\dfrac{7\pi }{6}+2k_2\pi:k_2\in Z\} $$

Where Z is the set of integers.

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From the unit circle definition of sine, we see that

$$\omega t = 2n\pi - \dfrac{\pi}{2} \pm \dfrac{\pi}{3}$$

and we can rephrase that as we wish.

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The sin$(x)$ function repeats itself every $2\pi$ times, so $\omega t=\frac{7\pi}6+2kπ$ and $\omega t=\frac{11\pi}{6}+2k\pi,\;$ where $k \in \mathbb Z$.

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Yes it is right but recall to add the $2k\pi$ term with $k\in \mathbb{Z}$, indeed we have

$$\sin(\omega t)=- \frac 1 2\iff \omega t=\dfrac{7\pi }{6}+2k\pi\,\lor\,\omega t=\dfrac{11\pi }{6}+2k\pi$$

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  • $\begingroup$ What if I don't recall to add $2k\pi$? $\endgroup$ – Mr. Maxwell Jul 7 '18 at 21:16
  • $\begingroup$ @Mr.Maxwell You can lost solutions. $\endgroup$ – user Jul 7 '18 at 21:17
  • $\begingroup$ However, isn't it enough to leave it up once we get $\dfrac{7\pi}{6}$ and $\dfrac{11\pi}{6}$? $\endgroup$ – Mr. Maxwell Jul 7 '18 at 21:18
  • $\begingroup$ @Mr.Maxwell Those are the solutions for $\omega t \in [0,2\pi]$ but in general solving trigonometric equations we need to consider a wider range. Take a look for example here math.stackexchange.com/q/2826222/505767 $\endgroup$ – user Jul 7 '18 at 21:20
  • $\begingroup$ @Mr.Maxwell - Enough is defined by who is asking the question. If the domain is not specified, then the question is poorly asked. $\endgroup$ – steven gregory Jul 7 '18 at 22:31

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