# Need help solving trig. equations.

• October 5th 2009, 03:37 PM
oryxncrake
Need help solving trig. equations.
How do I solve the following for y, where 0 is less than or equal to y and y is less than 360 degrees?

11sin^2y = 13 - 3sin^2y ?

Thank you.
• October 5th 2009, 04:14 PM
I-Think
$11sin^{2}y=13-3sin^{2}y$
$14sin^{2}y=13$
$sin^{2}y=\frac{13}{14}$
$siny=\sqrt{\frac{13}{14}}$
$y=sin^{-|}\sqrt{\frac{13}{14}}$
$y=74.5^o, 285.35^o$

Cheers!
• October 5th 2009, 05:58 PM
oryxncrake
Thanks! But how did you get 285.35 degrees?
• October 5th 2009, 07:13 PM
I-Think
Oops!
That should be $105.5^o$
Sorry for that mistake, misquoted an identity

To get the second answer, use the identity: $sin\theta=sin(180-\theta)$

The identity I misquoted was: $sin(360-\theta)=-sin\theta$