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![[Post New]](/templates/default/images/icon_minipost_new.gif) 6 Mar 2007 07:44:38 IST
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1. if tan  ,cos ,1/6sin are in gp, solve 3 .solve: cosec x cos2 x cos3x =1/4(0 <= x <= 2  ) 4.find the no. of solutions of tan x+secx =2 cosx(0 <= x <= 2 ) 5..if x n /2 and (cosx) sin^2x -3 sinx +2=1 find all such x
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sathya prabha girish |
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6 Mar 2007 12:48:27 IST
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First The Fourth Sum
sinx/cosx + 1/cosx = 2cosx
sinx+1/cosx = 2cosx
sinx + 1 = 2 cos^2 x
sinx + 1 = 2(1- sin^2 x)
sinx + 1 = 2 - 2 sin^2 x
2 sin^2 x + sin x - 1 =0
2 sin^2 x + 2sinx - sinx - 1 = 0
2sinx (sinx +1) - 1 (sinx + 1) =0
[2sinx - 1] [sinx +1] =0
Hence It Has 2 Solutions
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Sometimes One Dream Is Enough To Light Up The Entire October Sky....
First Year Mechanical Engineering
Veermata Jijabai Technological Institute |
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6 Mar 2007 13:08:56 IST
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sec x = ([sqrt]2 - 1) tanx
1/cos x = .... sin x /cos x
1 = ([sqrt]2 - 1) sin x
sinx = 1/([sqrt]2 - 1)
=[sqrt]2 + 1 / 2 - 1
(Rationalising)
BUT sin x value CANNOT EXCEED 1
THUS THIS EQN HAS NO REAL SOLUTION
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Sometimes One Dream Is Enough To Light Up The Entire October Sky....
First Year Mechanical Engineering
Veermata Jijabai Technological Institute |
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6 Mar 2007 13:13:05 IST
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A1> cos 2 =1/6cos cos =1/ [3] 6 so =2n(pi)+- cos -1(1/ [3] 6) |
~ANSHUMAN
I was born intellegent, education ruined me. |
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7 Mar 2007 08:06:06 IST
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but anshuman can u factorise( 6cos^3 +cos -1)??? |
sathya prabha girish |
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7 Mar 2007 08:15:18 IST
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hi nunoxic! what if the interval is between 0 and 2pi for the fourth sum
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sathya prabha girish |
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8 Mar 2007 11:19:25 IST
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to mr nunoxi
sum 4th has only 1 soln as rang is 0<=x<=2
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8 Mar 2007 12:20:01 IST
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sorry..i din see that.....thanks for pointing that out
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Sometimes One Dream Is Enough To Light Up The Entire October Sky....
First Year Mechanical Engineering
Veermata Jijabai Technological Institute |
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