Solve Trig Question: 14sin^2y-13=0 for y where 0<y<360

[Checkfate]

Hello,

I am running into an error and I can't quite figure out what I am doing wrong.

The question is "solve $$14\sin^{2}y-13=0$$ for y where 0<y<360.

So I factor it down to

$$14(sin(y)-\frac{\sqrt{182}}{14})(sin(y)+\frac{\sqrt{182}}{14})=0$$ (Thanks integral for showing me how to do this) :P

and thus get

$$y=74.5\deg or y=285.5\deg$$ by using arcsine, but I am missing solutions somehow. I checked my answer with my calc and it's wrong, I am missing 105.5^ and 254.5^... how though? Adding the period length of 360^ does absolutely nothing. It's probably a small detail that I am overlooking, can someone point me in the right direction? Thanks again.

 

[courtrigrad]

$$180 - 74.5 = 105.5$$

$$360 - 105.5 = 254.5$$