How do I solve the following for y, where 0 is less than or equal toyandyis less than 360 degrees?

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

Thank you.

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- Oct 5th 2009, 02:37 PMoryxncrakeNeed 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^2*y*= 13 - 3sin^2*y*?

Thank you. - Oct 5th 2009, 03:14 PMI-Think
$\displaystyle 11sin^{2}y=13-3sin^{2}y$

$\displaystyle 14sin^{2}y=13$

$\displaystyle sin^{2}y=\frac{13}{14}$

$\displaystyle siny=\sqrt{\frac{13}{14}}$

$\displaystyle y=sin^{-|}\sqrt{\frac{13}{14}}$

$\displaystyle y=74.5^o, 285.35^o$

Cheers! - Oct 5th 2009, 04:58 PMoryxncrake
Thanks! But how did you get 285.35 degrees?

- Oct 5th 2009, 06:13 PMI-Think
Oops!

That should be $\displaystyle 105.5^o$

Sorry for that mistake, misquoted an identity

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

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