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I don't seem to understand the concept of a negative domain when solving trigonometric equations on "another interval" For example:

Solve $\cos x=-\sqrt{3}/2$ given that the domain is $-\pi \le x\le \pi$.

I don't know what to do or understand the process at all! Please help, thank you!

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  • $\begingroup$ there's a nice interactive exhibit at mathopenref.com/trigterminalside.html that allows you to drag a point on the terminal side and see what angle (including negative ones) it results in. $\endgroup$ – John Joy Nov 7 '14 at 16:10
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$$\cos x=-\frac{\sqrt3}2=-\cos\frac\pi3=\cos\left(\pi-\frac\pi3\right)$$

$$x=2m\pi\pm\left(\pi-\frac\pi3\right)$$ where $m$ is any integer

Now find $m$ such that $-\pi\le x\le\pi$

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Simple. $x = -pi/6 $ will do nicely. To see this, note that cos is periodic with period $\pi$. This means that $x+\pi = x$ for all $x$. Note however this does not mean that $x=x-\pi$ for all $x$. Anyway, you can find the answer in the interval $0<x<\pi$, then just translate the interval.

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