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![[Post New]](/templates/default/images/icon_minipost_new.gif) 23 Mar 2008 23:32:58 IST
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solve the following trigonometric equations...
1. sqrt(3)*(2-cosx) + 4sin2x = sinx
2. 20cos^2(x) = 5 +sinx + sqrt(3)cosx
my only guess is that
sinx + sqrt(3)*cosx
can be written as
2sin(x+pi/3)
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23 Mar 2008 23:43:05 IST
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2root3 -root3cosx + 4sin2x = sinx
=> root3 + 2sin2x = sinx/2 + root3cosx/2
=> root3 + 2sin2x = sin(30+x)
=> 2(sin60 + sin2x) = sin (30 +x)
=> 4sin (30+x)cos(30-x) = sin(30+x)
=> sin(30+x)(1 - 4cos(30-x)) =0
=> sin(30+x) =0 OR cos(30-x) = 1/4
After this is easy but hectic to type.
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Let us learn to dream, gentlemen, and then perhaps we shall learn the truth.
- August Kekule |
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24 Mar 2008 00:08:06 IST
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thanks... what about the second problem??
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24 Mar 2008 09:21:25 IST
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20 cos^2x=5+sinx+  3cosx 20cos^2x-5=2cosx-  /3 5(4cos^2x-1)=2cosx- 3 5(2cos^x+cos2x)=2cosx- /3 LHS min=5 rhs max2 2cos^x+cos2x=o and cosx- /3=0 solve get ans
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24 Mar 2008 12:34:06 IST
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^^
pi/6 instead of pi/3 dude.
Rest is allright.Nice
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Let us learn to dream, gentlemen, and then perhaps we shall learn the truth.
- August Kekule |
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