Stacy Rae

Posted on Monday, 21 April, 2003 - 11:21 pm:

I cannot seem to get my head around this question at all so have given in and asking for help!!

Q) $5 \sin^{2} y = 3$ find all solutions for y in degrees (0-360) and radians $(0 - 2 π)$ .

I have worked this out so far by changing the $\sin^{2}$ to $1 - \cos^{2}$ and so have come up with $5 (1 - \cos^{2} y)$ ; expanding out the brackets gives me $5 - 5 \cos^{2} y = 3$ and putting into a quadratic:- $5 - 5 \cos^{2} - 3 = 0$ . Therefore $2 - 5 \cos^{2} y = 0$ .

It's here that my brain has given up. Am I on the right track at all? What's the best way to proceed on please.

thanx stacy :-)

p.s This is for my hnc and the last time I did anything like this was about 8 years ago at college of fe so simple would be great please!!

Colin Prue

Posted on Tuesday, 22 April, 2003 - 01:10 am:

\sin^{2} y = \frac{2}{5}

\Rightarrow \sin y = \sqrt{\frac{2}{5}}

\Rightarrow y = \sin^{- 1} (\sqrt{\frac{2}{5}})

I don't think there is a nice answer for this, just whack it in your calculator!

Stacy Rae

Posted on Tuesday, 22 April, 2003 - 08:45 am:

oh dear was it really that simple!!
thanx for your help
:-)

David Loeffler

Posted on Tuesday, 22 April, 2003 - 09:11 am:

If you're looking for all solutions, remember that there is a second solution, 180° - this.

David

Andre Rzym

Posted on Tuesday, 22 April, 2003 - 09:21 am:

Don't forget that a square root has two values (positive and negative), therefore there are four solutions to your equation in the range 0..2pi .

Andre

Stacy Rae

Posted on Tuesday, 22 April, 2003 - 01:22 pm:

thanks all for your help, have worked out solutions in all four quadrants, it was mainly what to do with a power against the sin function, i was working on the principle that to remove it you had to cx the sin function as well!
thanx again
stacy