• Mar 16th 2006, 03:57 PM
hunisukl
I can not figure out how to work this problem, can any one help???

sin(x) = -1/2

I would appreciate any help. Thanks!
• Mar 16th 2006, 06:58 PM
ThePerfectHacker
Quote:

Originally Posted by hunisukl
I can not figure out how to work this problem, can any one help???

sin(x) = -1/2

I would appreciate any help. Thanks!

I think what is bothering you is that not that you can solve this equation for some x but to solve it for all x.

You begin by finding one solution of,
$\sin x=-1/2$
You can do this by using arcsine, but you should know what this angle is,
$x=\frac{7\pi}{6}$
and
$x=\frac{11\pi}{6}$
These are all the non-coterminal angles, thus, all solutions are,
$x=\frac{7\pi}{6}+2\pi k$
$x=\frac{11\pi}{6}+2\pi k$

The problem which many people have is finding what the angle has to be. Notice that if the problem was $\sin x=1/2$ then the angle is $\frac{\pi}{6}$ you should have this memorized, but the problem asks for $\sin x=-\frac{1}{2}$ thus, one such solution is, $-\frac{\pi}{6}$. Now find where this is located the on circle and in which quadrant. This is located in 4th quadrant producing angle of $\frac{11\pi}{6}$. How do you know what this angle is? Well, you now it is $\frac{\pi}{6}$ backwards and the full circle has $2\pi$ thus if you subtract them properly you get $\frac{11\pi}{6}$. To find the other non-coterminal solution you now since it is negative it is in the 3rd quadrant. Thus, it is $\frac{\pi}{6}$ below the $\pi$ axis. Thus, a total of $\frac{7\pi}{6}$ if you add them properly. Now since adding $2\pi$ or subtracting it does not change the sines values they are all solutions thus, this is what $2\pi$ is doing there.
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There is a much easier way. Look at this:
$\sin x=a$ for $-1\leq a\leq 1$
Then ALL the solutions for $x$ are,
$x=(-1)^k\sin^{-1}(a)+\pi k$
Where $\sin^{-1}(a)$ is the inverse sine function.

Thus, in your problem you get,
$x=(-1)^k\sin^{-1}(-1/2)+\pi k$
But $\sin (-1/2)=-\frac{\pi}{6}$
Thus,
$x=(-1)^k\frac{-\pi}{6}+\pi k=(-1)^{k+1}\frac{\pi}{6}+\pi k$
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You might be thinking then what is the difference between the first solution I gave and this one? The answer is there is no difference, the second one is a fancier way of expressing all the solutions for $x$. In fact, if were to write them out you will see you get the exact same thing.