# Thread: Need help solving trig. equations.

1. ## 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.

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

Cheers!

3. Thanks! But how did you get 285.35 degrees?

4. 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: $\displaystyle sin(360-\theta)=-sin\theta$