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Trignometry

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27 Feb 2008 18:56:58 IST

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541

Trig Eqn

None

Find x

R such that sin³x+cos³x =

2 sinx cosx.

And if you have the time x

C also!

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Comments (7)

Akhil

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27 Feb 2008 19:12:29 IST

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2n(pi)+pi/4

(sin x+cos x)(1-sin x cos x)=

2 sin x cos x
is the equation
on dividing by RHS and rearranging we get
secx+cosec x-(sin x+cos x)=

2
from this i am getting that all solns lie in first quadrant...
and for pi/4

Hari Shankar

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27 Feb 2008 19:13:44 IST

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Many thanks for using solution by inspection.

There are some more solutions, pls find them too.

The Sniper

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27 Feb 2008 20:03:30 IST

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$sin^3x+cos^3x=\sqrt{2}sinxcosx\\ \\ tan^3x+1=\sqrt{2}tanx\sqrt{1+tan^2x}\\ \\ \text{let tanx=m }\\ \\ m^3+1=\sqrt{2}m\sqrt{1+m^2}\\ \\ =m^6-2m^4+2m^3-2m^2+1=0\\ \\ =(m-1)^2(m^4+2m^3+m^2+2m+1)=0\\ \\ m=1,-1.86,-0.47\\ \\ \theta=2n\pi+\pi/4,2n\pi+(-1.077),2n\pi+(-.439)$

anchit saini

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27 Feb 2008 20:07:33 IST

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the eqn is

(sin x+cos x)(1-sin x cos x)=(root2) sin x cos x

take sin x + cos x=t

therefore
1 + 2 sinx cos x= t^2
(t^2-1)/2=sinx cos x

hence the eqn becomes
t(3-t^2)/2 =(root 2)(t^2-1)/2
therefore
3t-t^3=root2*t^2-root2

t^3+ root 2* t^2 - 3t -root(2)=0 ----------1

on inspection
t=root 2 -----------2
is already a root
which gives one value of
x=pi/4
rest can be found using
1 and 2

final eqn is--
(t-root2)(t^2 + 2root 2t + 1)=0
which gives other values of
t=-root2+1
, -root2-1

sinjan jana

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27 Feb 2008 22:11:38 IST

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given eq. can be converted into the eq. by putting cosx-sinx=t
2t^3-t^2-3t+1=0
where t=sin(x-PI/4)/sqrt2
so
x=PI/4 +nPI+(-1)^n*arcsin(sqrt 2*t)
where t is the root of above eq. with mod t <=1/sqrt2
and n is integer

Hari Shankar

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28 Feb 2008 09:13:48 IST

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good job anchit. This was the soln I was looking for which gives the answers neatly.

anchit saini

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28 Feb 2008 09:34:24 IST

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thank u sir

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