Trigonometry
posted by Blair .
4+29sin=12cos^2
How do you determine this

If your expression mean:
4+29sin(x)=12cos^2(x)
then go on:
wolframalpha dot com
When page be open in rectangle type:
solve 4+29sin(x)=12cos^2(x)
and click option =
After few seconds when you see result click option: Show steps 
Remark:
cos ^ 2 ( x ) = 1  sin ^ 2 ( x )
12 cos ^ 2 ( x )= 12 [ 1  sin ^ 2 ( x ) ] =
12  12 sin ^ 2 ( x )
29 sin ( x )  12 cos ^ 2 ( x ) + 4 =
29 sin ( x )  [ 12  12 sin ^ 2 ( x ) ] + 4 =
29 sin ( x )  12 + 12 sin ^ 2 ( x ) + 4
= 29 sin ( x )  8 + 12 sin ^ 2 ( x ) =
12 sin ^ 2 ( x ) + 29 sin ( x )  8 
When I solved this out I got x=32/12 and x=1/4. Is this right?

Not x = .... but rather sinx = 32/12 or sinx = 1/4
Picking up from where Bosnian left off
12sin^2 x+ 29sinx  8 = 0
(4sinx  1)(3sinx + 8) = 0
sinx = 1/4 or sinx = 8/3, but sinx has to be between 1 and 1, so the last part is undefined
sinx = 1/4, so
x = 14.48° or 165.52°
if you want radians, set your calculator to RAD
and find arcsin (.25)
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