# Trigonometry

posted by Blair

4+29sin=12cos^2

How do you determine this

1. Bosnian

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

2. Bosnian

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

3. Blair

When I solved this out I got x=-32/12 and x=1/4. Is this right?

4. Reiny

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°

and find arcsin (.25)

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