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Another trig question

  • Thread starter Checkfate
  • Start date
  • #1
148
0
Hello,

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

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

So I factor it down to

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

and thus get

[tex]y=74.5\deg or y=285.5\deg[/tex] 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.
 

Answers and Replies

  • #2
1,235
1
[tex] 180 - 74.5 = 105.5 [/tex]

[tex] 360 - 105.5 = 254.5 [/tex]
 

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