Originally Posted by

**Archie Meade** Hi sigma1,

the quadratic equation allowed you to discover the 2 possible values for $\displaystyle sinx$

Now you need to find the valid values of the angle x.

$\displaystyle sinx$ may be conveniently thought of as the vertical co-ordinate

of a point on a circle centred at (0,0) with radius 1.

Then you can see, using a sketch that a vertical co-ordinate of $\displaystyle \frac{1}{3}$

gives 2 possible angles, $\displaystyle x=sin^{-1}\left(\frac{1}{3}\right)$

and $\displaystyle 180^o-sin^{-1}\left(\frac{1}{3}\right)$

For $\displaystyle sinx=-\frac{1}{2}$ on the vertical axis,

you can calculate the acute angle $\displaystyle sin^{-1}\left(\frac{1}{2}\right)$

to get the acute angle the point on the circle makes under the x-axis.

Then subtract this from 360 degrees and add it to 180 degrees.

You really need to understand that procedure