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General Solution (1 Viewer)

McLake

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A while ago somone asked for an example of a "general solution" question with two answers.

Well, here's an example:

sin2x = cosx
 

Dumbarse

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sin2x = cosx

2 sinx cosx - cosx = 0

cosx (2sinx - 1) =0

cosx = 0 sinx = 1/2

x = 2npi +- cos^-1 (0) 0r x = npi + (-1)^n *sin ^-1 (1/2)

x = 2npi +- pi/2 x = npi + (-1)^n* pi/6


how bout something harder??

give the general solution for

1) cos 5x + cos 7x = cos x + cos 3x wait thats 4 unit

how bout

2) Tan 2x = 2 sinx
 

McLake

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Does your 4U Q use De Movories (or however it's spelt)?

Q2)

tan2x = 2sinx
2tanx/(1 - tan^2x) = 2sinx
2tanx = 2sinx - 2sinx(tan^2x)
2tanx = 2sinx - 2(sin^3x/cos^2x)
2tanx = 2sinx - [2sinx(1 - cos^2x)/cos^2x]
2tanx = 2sinx - 2(sinx/cos^2x) + 2sinx
2tanx = 4sinx - 2(sinx/cos^2x)

* cos^2x

2sinxcosx = 4sinxcos^2x - 2sinx

/ sinx (sinx != 0)

2cosx = 4cos^2x - 2
4cos^2x - 2cosx - 2 = 0

QUADRATIC FORMULA

cosx = 1 or -1/2
x = cos^-1(1) AND x = cos^-1(-1/2)
x = 0 and 2/3(pi)

Check sinx = 0
x = 0
Already used but genral formula needs to be applied ...

x = {n(pi) , 2n(pi) +/- 2/3(pi)}

NB: n(pi) covers cosx = 0 AND sinx = 0
 
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Jellymonsta

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Originally posted by Dumbarse

x = 2npi +- cos^-1 (0) 0r x = npi + (-1)^n *sin ^-1 (1/2)

x = 2npi +- pi/2 x = npi + (-1)^n* pi/6
Eh? Where does this come from? Or is it 4U stuff that would make my ears bleed?
 

DA

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Does your 4U Q use De Movories (or however it's spelt)?
The 4U question can be done by writing cos 7x + cos 5x = cos (6x+x) + cos (6x-x) and cos x + cos 3x = cos (2x - x) + cos (2x + x), and then simplifying by using cos (A+B) + cos (A-B) = 2cos A cos B.
 

McLake

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Originally posted by Jellymonsta

Eh? Where does this come from? Or is it 4U stuff that would make my ears bleed?
That's just general solution ...
 

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