Hey I'm not sure how you solve this. Calculate to the nearest degree the values of x when sin x = 0.2, for -180degrees < x < 180degrees.

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- Apr 23rd 2006, 01:34 PMdigideensSin Graph Help
Hey I'm not sure how you solve this. Calculate to the nearest degree the values of x when sin x = 0.2, for -180degrees < x < 180degrees.

- Apr 23rd 2006, 01:47 PMCaptainBlackQuote:

Originally Posted by**digideens**

sin function on a calculator to do this. It can be done using Newton Raphson

or a number of other techniques.

There is also the problem that there are two solutions in the specified range.

RonL - Apr 23rd 2006, 01:48 PMThePerfectHackerQuote:

Originally Posted by**digideens**

Thus, for

$\displaystyle \sin x=.2$

You find, (have calculator in degree mode).

$\displaystyle x=\sin^{-1}(.2)\approx 11.54^o$

But that is still not it. Because the sine is also positive in the second Quadrant. In the second quadrant the value is,

$\displaystyle 180^o-11.54^o\approx 168.46^o$ - Apr 23rd 2006, 01:51 PMdigideens
Thanks for the quick replies guys, I got it now.