general trig solutions Watch

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ryan750
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#1
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#1
sin4x = cos2x

i seem to have got up to thias point:

cos2x(2sin2x -1) = 0

but from here my answers to the book dont match up

it has one of the soltuions as kpi -pi/4

i think i have a 2 under the kpi. which is right?
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ryan750
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#2
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its ok i think i realise now
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ryan750
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#3
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the other soution doesn't quite match up to mine either. i got mine from 2sin2x-1 = 0

wot would be the general solution for this and why?
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Feria
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#4
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for cos2x(2sin2x -1) = 0

you not only have to consider when (2sin2x -1) = 0, but also if cos2x = 0 because either one could be zero to satisfy the equation, and its not nessacarily the second half
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ryan750
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#5
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(Original post by Feria)
for cos2x(2sin2x -1) = 0

you not only have to consider when (2sin2x -1) = 0, but also if cos2x = 0 because either one could be zero to satisfy the equation, and its not nessacarily the second half
i have done the cos2x = 0 bit. but my solution to the other half doesnt match to the books answer. its almost there - but there seems to be sum extra coeffiecient.
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ryan750
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#6
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#6
i get two solution for sumthing right.

when sinx = 1/2
x = 2npi + pi/6, 2npi + 5pi/6 - correct?

but my book seems to have kind of put these together to make 2npi/3 + pi/6

altho this seems to be for sinx = +/- 1/2
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Feria
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#7
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are you still considering that there is more than one value for sin2x=0? at sinx=0, x=0, pi, 2pi etc... so at sin2x=0, x=0, pi/2, pi etc... at cosx=0, x=pi/2, 3pi/2, 5pi/2... so at cos2x=0, x=pi/4, 3pi/4, 5pi/4... (or 1/2pi - pi/4, 1pi-pi/4, 3/2pi-pi/4) - which is your kpi-pi/4
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ryan750
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#8
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#8
Ive found the problem - the textbook i use makes use of the fact that sin X = cos(pi/2 - X)

simple i know - but it makes all the difference - YAYYYYYYYYYYYYY!!!!!!!!!!!!!!
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RichE
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#9
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#9
(Original post by ryan750)
sin4x = cos2x
i seem to have got up to thias point:
cos2x(2sin2x -1) = 0
Either

cos(2x) = 0
2x = (n+1/2)pi
x = npi/2 + pi/4

(NB this is the same set of solutions as npi/2 - pi/4)

or

sin2x = 1/2
2x = pi/6+2npi or 5pi/6 + 2npi
x = pi/12 + npi or 5pi/12 + npi
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ryan750
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#10
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#10
what is the general solution for
(cosx + sinx)^2 = 3/2
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Gaz031
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#11
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(Original post by ryan750)
what is the general solution for
(cosx + sinx)^2 = 3/2
1+2sinxcosx=3/2
sin2x=1/2
sin(2x+2kpi)=1/2
It's easy to proceed from there.
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ryan750
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#12
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#12
thanks - i think i might actually be getting used to this stupid topic - its been so hard to get my head round it for sum reason. :tsr:
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ryan750
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#13
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#13
last question on this exercise:

i have one solution but dont know how to get the other solution:

find the general solution of sqrt3cos2x + sin2x = 2sinx

i get;
cos(2x - pi/6) = cos(pi/2 - x)

(2x - pi/6) = (pi/2 - x)

3x = 2pi/3 + 2npi
x = 2pi/9 + 2npi/3

thats one correct soltuion
the other in the answers is 2npi - pi/3

how do i get this one?
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RichE
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#14
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#14
(Original post by ryan750)
last question on this exercise:

i have one solution but dont know how to get the other solution:

find the general solution of sqrt3cos2x + sin2x = 2sinx

i get;
cos(2x - pi/6) = 2cos(pi/2 - x)

(2x - pi/6) = (pi/2 - x)
well that line isn't valid you have a 2 in front of the RHS

EDIT - ah ignore that, you seem to have edited the working
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RichE
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#15
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#15
I already mentioned how you could do

cosP = cosQ

equations by noting

cos(A+B) - cos(A-B) = -2sinAsinB
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ryan750
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#16
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#16
how do i use that here?
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RichE
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#17
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#17
solve

A+B = 2x -pi/6

and

A-B = pi/2 -x
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RichE
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#18
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#18
actually because cos is even it's even easier than this

cosP = cosQ

means

P = Q + 2npi

or

P = -Q + 2npi
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ryan750
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#19
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#19
yep - have A = x/2 + pi/6 and B = -3x/2 + pi/3
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ryan750
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#20
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#20
(Original post by RichE)
actually because cos is even it's even easier than this

cosP = cosQ

means

P = Q + 2npi

or

P = -Q + 2npi
yes i did it this way to get one correct answer using P = Q + 2npi

but i dont think the answer in th bak of the book matches to9 my ansa wen i use P = -Q + 2npi
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