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![[Post New]](/templates/default/images/icon_minipost_new.gif) 6 Apr 2008 20:10:02 IST
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2sin-1x * cos-1x = 3cos-1(1/ x) *sin-1(1/x
if anyone could solve it please respond... will rate
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6 Apr 2008 21:43:26 IST
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my goodness..!! ..i can't read questions properly..crosses all over ..
wat sign did u use below 1 on r.h.s..
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Diamonds r formed under greatest pressures..
so r the champs.
Kriteesh..
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7 Apr 2008 01:07:58 IST
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thats square root of x...
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7 Apr 2008 02:05:33 IST
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since 1/x shud lie between 1 and -1,x sud be greater than or equal to 1..and -1 on that side
[considering the domain of cos and sin inv]
but in the left hand side x cannot take value than1
so the only ways is when x = 1which doesnt satisfy..
hence no solution
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7 Apr 2008 06:14:21 IST
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2sin-1x * cos-1x = 3cos-1(1/ sqrt x) *sin-1(1/sqrt x)
since domain on LHS is x between -1 to 1.
and domain on RHS is x greater than equal to 1.
so boundary value is 1.
and 1 satisfies the equation , as cos -1( 1) is 0 in both LHS and RHS.
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* Gaurav Ragtah ( aka Artemis Fowl )
* Agent 'G' [sniper] - SD-6 (Alliance of Twelve)
* Your friendly neighborhood spideyunlimited |
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