Need help solving trig. equations.

**oryxncrake** · October 5th 2009, 02:37 PM

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.

**I-Think** · October 5th 2009, 03:14 PM

$11sin^{2}y=13-3sin^{2}y$
$14sin^{2}y=13$
$sin^{2}y=\frac{13}{14}$
$siny=\sqrt{\frac{13}{14}}$
$y=sin^{-|}\sqrt{\frac{13}{14}}$
$y=74.5^o, 285.35^o$

Cheers!

**oryxncrake** · October 5th 2009, 04:58 PM

Thanks! But how did you get 285.35 degrees?

**I-Think** · October 5th 2009, 06:13 PM

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: $sin(360-\theta)=-sin\theta$

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