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Sinx+cosx=5, tanx+secx=3 then sinx=?

Sinx+cosx=5,tanx+secx=3

sinx=?

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Tanx= sinx/cosx,
secx=1/cosx
1+sinx/cosx=3
1+sinx=3cosx---------->1
cosx=5-sinx------------->2

sub 2 in 1
1+sinx=3(5-sinx)
1+sinx=15-3sinx
1+4sinx=15
4sinx=14
sinx=14/4...
this is the solution..

Posted on Jul 23, 2008

Sin180-x

Posted on Oct 22, 2008

5cos A +12sinA=4

Posted on Aug 30, 2008

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1 Answer

How can I solve Sinx+cosx=7/5?


Square each side

Sin ^2 (x) + cos ^2 (x) +2 sin (x)cos (x) = 49/25

1 + 2sin (x)cos (x) = 1.960

sin (2x) = 0.960

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Use the identity tan(x/2)=sinx/1+cosx to solve for the value of tan45 degrees


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SinX-cosX=4 how can this be solved?


I gues this problem has no solution because both
SinX and cosX should be between -1 and +1 so it's not possible that value of SinX - cosX should be 4

Dec 12, 2010 | ValuSoft Bible Collection (10281) for PC

1 Answer

Cos2x + 3 = 5cosx


the solution you got is correct it is

cos X =2 and cos X =1/2


but we know maximum value of cosX is 1.so we discard the solution cos X =2


so only solution is cos X=1/2

and X = 60 degrees

hope you r satisfied.please rate the solution high.thank you

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1 Answer

Solve for X . tanx + secx = 3


Hint: split it up into 
int sec^2(x) dx + int e^sin(x) cos(x) dx
The first one is standard, and the second one is a straightforward u-substitution.

Jul 20, 2009 | ValuSoft Bible Collection (10281) for PC

1 Answer

(1+cotx-cosecx)(1+tanx+secx)=2


I shall attempt :D
1) cosec A + cot A = 3
we know that (cot A)^2 + 1 = (cosec A)^2
Hence, (cosec A)^2 - (cot A)^2 = 1
thus, (cosec A + cot A) (cosec A - cot A) = 1
3 (cosec A - cot A) = 1
(cosec A - cot A) = 1/3

(cosec A - cot A) = 1/3
(cosec A + cot A) = 3
Summing them, 2 cosec A = 3 1/3
cosec A = 6 2/3 = 5/3
sin A = 0.15
Thus, cos A = sqrt (1 - (sin A)^2) = 0.989


2) Prove that (1+tan x - sec x)(1 + cot x + cosec x) =2
expand
LHS= 1 + cot x + cosec x + tan x + 1 + tan x cosec x - sec x - sec x cot x - sec x cosec x
We can calculate that
tan x cosec x = sec x (since tan x = sin x / cos x)
sec x cot x = cosec x
so the above is
LHS = 1 + cot x + cosec x + tan x + 1 + sec x - sec x - cosec x - sec x cosec x
LHS = 2 + cot x + tan x - sec x cosec x
LHS = 2 + cos x / sin x + sin x / cos x - 1 / (sin x cos x)
LHS = 2 + [{cos x}^2 + {sin x}^2 - 1] / (sin x cos x)
LHS = 2 (proved)

May 12, 2009 | ValuSoft Bible Collection (10281) for PC

1 Answer

(sin-cos)(sin+cos)


(sinX-cosX)(sinX+cosX) =(sin^2 - Cos^2X)
=Sin^2X - (1-Sin^2X)
=Sin^2X -1 + Sin^2X
=2Sin^2X -1

Zulfikar Ali
ali_zulfikar@yahoo.com
9899780221

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Wat is the solution for sinx-cosx=2 ?


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