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How could I solve the following trigonometric equation?

$$\sin(\omega t)=- \frac 1 2$$

Take the inverse sine of both sides:

$$\omega t=\arcsin\left( -\frac 1 2 \right)$$

$$t=\frac 1 \omega \arcsin\left( -\frac 1 2 \right) =- \frac {6\pi}{\omega}$$ and then I don't know.

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see my answer:

when $\sin\left(2k\pi- \frac{\pi}{6}\right)=-1/2$ then , $$\sin(\omega t)=-\frac 12=\sin\left(2k\pi- \frac{\pi}{6}\right)$$ $$\omega t=2k\pi- \frac{\pi}{6}$$ $$t=\frac{(12k-1)\pi}{6\omega}$$ when $\sin\left(2k\pi+ \frac{7\pi}{6}\right)=-1/2$ then , $$\sin(\omega t)=-\frac 12=\sin\left(2k\pi+ \frac{7\pi}{6}\right)$$ $$\omega t=2k\pi+ \frac{7\pi}{6}$$ $$t=\frac{(12k+7)\pi}{6\omega}$$ take, $k=0, \pm1, \pm2, \pm3, \ldots$

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    $\begingroup$ Much better +1. $\endgroup$ – imranfat Oct 18 '16 at 17:24
  • $\begingroup$ Hi @Bhaskara-III, why you wrote $\sin\left(2k\pi- \frac{\pi}{6}\right)=-1/2$ and $\sin\left(2k\pi+ \frac{7\pi}{6}\right)=-1/2$? $\endgroup$ – Gennaro Arguzzi Oct 18 '16 at 17:36
  • $\begingroup$ because general solution to the equation $\sin \theta=-\frac 12$ is $$\theta=2k\pi-\frac{\pi}{6}\ \text{or}\ \theta=\ 2k\pi+\frac{7\pi}{6}$$ $\endgroup$ – Bhaskara-III Oct 18 '16 at 17:40
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    $\begingroup$ Thank you @Bhaskara-III :) $\endgroup$ – Gennaro Arguzzi Oct 18 '16 at 17:49
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First observe for what values does sine value equals -0.5

Then for those values equate to, then you will get answer for Omega as well as time period.

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    $\begingroup$ This is a comment, not an answer... $\endgroup$ – imranfat Oct 18 '16 at 17:17
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Given: $$\sin(\omega t)=- \frac 1 2$$

First: Rewrite the right hand side as a sine: $$\sin(\omega t)=sin\frac{-\pi}{6}$$

There are two branches that follow from this: $\omega t=\frac{-\pi}{6}+2k\pi$ and $\omega t=\pi-\frac{-\pi}{6}=\frac{7\pi}{6}+2k\pi$. This simplifies too $t=\frac{-\pi}{6\omega}+\frac{2k\pi}{\omega}$ or $t=\frac{7\pi}{6\omega}+\frac{2k\pi}{\omega}$ for all integer $k$'s

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