the question is:
Solve sinx=cos(x+30) for values of x 0<x<360.
So far I've done:
sin x = cosxcos30  sinxsin30
Now I'm not sure what to do can anyone help?

meiming8
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 06022010 14:29

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 06022010 14:34
cos30 and sin30 are just numbers (constants).
Factorise the sinx and use a C2 identity. 
paronomase
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 06022010 15:06
In other words, you have
sin(x)=cos(x+30)
<=> sin(x)=cos(x)cos(30)sin(x)sin(30)
<=> sin(x) = [sqrt3cos(x)/2]  [sin(x)/2]
<=> 3sin(x)/2 = sqrt 3 cos(x) / 2
<=> 3 sin(x)=sqrt3 cos(x)
<=> tan(x)= (1/sqrt3)
<=> x = pi/6 [pi] 
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 06022010 15:19
(Original post by paronomase)
<=> tan(x)= (1/sqrt3)
<=> x = pi/6 [pi] 
paronomase
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 06022010 15:48
In formal logic, "<=>" is not implication, but equivalence. It is possible that I've made an error, but where would that be?
The implication sign is "=>". You may want to check this http://en.wikipedia.org/wiki/Logical_implication 
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 06022010 16:49
(Original post by paronomase)
In other words, you have
<=> 3 sin(x)=sqrt3 cos(x)
<=> tan(x)= (1/sqrt3)
<=> x = pi/6 [pi] 
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 06022010 16:56
3 sin(x) = sqrt 3 cos (x)
<=> 3 sin(x) / cos (x) = sqrt 3 (for x different of pi/2 [pi] otherwise you cannot divise by 0)
<=> 3 tan (x) = sqrt 3 since tan(x) = sin(x) /cos(x)
<=> tan (x) = sqrt 3 / 3
<=> tan (x) = 1 / sqrt 3 
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 06022010 17:20
(Original post by meiming8)
the question is:
Solve sinx=cos(x+30) for values of x 0<x<360.
So far I've done:
sin x = cosxcos30  sinxsin30
Now I'm not sure what to do can anyone help?
sinx = sqrt(3)cosx / 2  sinx / 2
sinx = sqrt(3)cosxsinx / 2
2sinx = sqrt(3)cosx  sinx
3sinx = sqrt(3)cosx
3tanx = sqrt(3)
tanx = sqrt(3)/3
tanx = sqrt(3)*sqrt(3) / 3*sqrt(3)
tanx = 3 / 3sqrt(3)
tanx = 1/sqrt3
x = 30 or 210 
paronomase
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 06022010 17:29
(Original post by innerhollow)
sinx = cosxcos30  sinxsin30
sinx = sqrt(3)cosx / 2  sinx / 2
sinx = sqrt(3)cosxsinx / 2
2sinx = sqrt(3)cosx  sinx
3sinx = sqrt(3)cosx
3tanx = sqrt(3)
tanx = sqrt(3)/3
tanx = sqrt(3)*sqrt(3) / 3*sqrt(3)
tanx = 3 / 3sqrt(3)
tanx = 1/sqrt3
x = 30 or 210
x=30 or x=210 and x=pi/6 [pi] is exactly the same thing, in the case where 0<x<2pi...(it should be three bars on the "=" sign though, but I did not want to use latex for one symbol) 
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 06022010 17:40
(Original post by paronomase)
In formal logic, "<=>" is not implication, but equivalence. It is possible that I've made an error, but where would that be?
The implication sign is "=>". You may want to check this http://en.wikipedia.org/wiki/Logical_implication
<==> means ==> and <== where the former means "P implies Q" and the latter means "P is implied by Q". The short form of writing ==> and <== is <==>, so it's an implication meaning "if and only if". 
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 06022010 17:44
(Original post by paronomase)
x=30 or x=210 and x=pi/6 [pi] is exactly the same thing, in the case where 0<x<2pi...(it should be three bars on the "=" sign though, but I did not want to use latex for one symbol)
Does anyone know how to enter line breaks into latex without using lots of tags by the way? If you only use one latex tag, it puts everything on the one line.
(Original post by Swayum)
It is an implication; two implications.
<==> means ==> and <== where the former means "P implies Q" and the latter means "P is implied by Q". The short form of writing ==> and <== is <==>, so it's an implication meaning "if and only if".
===> is only a oneway implication.Last edited by innerhollow; 06022010 at 18:03. 
paronomase
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 06022010 17:57
(Original post by Swayum)
It is an implication; two implications.
<==> means ==> and <== where the former means "P implies Q" and the latter means "P is implied by Q". The short form of writing ==> and <== is <==>, so it's an implication meaning "if and only if".
"<=>" is "<=" AND "=>" so it is TWO implications so it cannot be ONE. Obvious no?
If P <=> Q, saying that P => Q is only a partial affirmation, and therefore you are wrong (unless you do not want to completely describe the relations between the two expressions)
Therefore, what is wrong with using " <=> " in my case?
I will not copy everything again, but what is the mistake in writing:
sin(x)=cos(x+30) <=> sin(x)= cos(x)cos(30)sin(x)sin(30)
Indeed, cos (x+30 ) = cos(x)cos(30)  sin(x)sin(30)
To rewrite it more simply,
If you have A=B, and
[1] A = C => B = C
[2] B = C => A = C
Then, with [1] and [2], you can write A=C <=> B=C
Just replace A by sin (x) and C by cos (x+30) and B by .... and you have the result 
paronomase
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 06022010 18:07
Yeah basically I just wrote it in radians and using modular arithmetics because it is easier. Have you done it yet? To explain a little bit...
I wrote x=pi/6 [pi] (read x is congruent to pi/6 modulus pi)
This means that x= pi/6 + kpi, where k is an integer
I precised that x belongs to [0,2pi] therefore k=1 or k=0
Therefore, x = pi/6 or x=pi/6 + pi = 7pi/6
Another example: if you want to solve cos (x) = 1/2,you can write either
(1) x = 60° or x=300°
(2) x=pi/3 or x=2pi pi/3 = 5pi/3
(3) x= pi/3 [2pi] or x=5pi/3 [2pi]
The advantage of (3) is that it tells you that basically there is an infinite number of solutions, if you add 2pi each time. 
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 06022010 18:22
(Original post by paronomase)
Yeah basically I just wrote it in radians and using modular arithmetics because it is easier. Have you done it yet? To explain a little bit...
I wrote x=pi/6 [pi] (read x is congruent to pi/6 modulus pi)
This means that x= pi/6 + kpi, where k is an integer
I precised that x belongs to [0,2pi] therefore k=1 or k=0
Therefore, x = pi/6 or x=pi/6 + pi = 7pi/6
Another example: if you want to solve cos (x) = 1/2,you can write either
(1) x = 60° or x=300°
(2) x=pi/3 or x=2pi pi/3 = 5pi/3
(3) x= pi/3 [2pi] or x=5pi/3 [2pi]
The advantage of (3) is that it tells you that basically there is an infinite number of solutions, if you add 2pi each time.
But wait? Surely that [pi] thing only works for tan, because for sin/cos there is no constant you can add to generate the next possible solution ad infinitum. For example, the first line is what you said.
...Add 2pi
...Add 2pi
But it doesn't work, because you're missing out solutions 
meiming8
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 06022010 18:28
(Original post by innerhollow)
Oh I see. I wondered why you kept writing a pi in square brackets after.
But wait? Surely that [pi] thing only works for tan, because for sin/cos there is no constant you can add to generate the next possible solution ad infinitum. For example, the first line is what you said.
...Add 2pi
...Add 2pi
But it doesn't work, because you're missing out solutions 
paronomase
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 06022010 19:18
Yes, it does not always work
If you are looking at an interval [0,2pi] then if you have an equation of the form cos(x)=k or sin(x)=k you solve it and you get X1 = ... and X2 = ...
Here, you cannot add a constant, as you said.
For cos(x)=k, you have two solutions: one that is X1 and the other that is X2=X1 (for example cos(x)=1/2 <=> x=pi/3 or x=pi/3
For sin(x)=k, you have two solutions X1 and X2=piX1
(for example, sin(x) = 1/2 <=> x= pi/6 or x= pipi/6=5pi/6)
Therefore, when solving trigonometric equations, one should be careful.
Modular arithmetics are really useful, but in trigonometry, the main use is just to show that there is an infinity of solutions for trig equations.
Here is another example: you have to solve cos(x)sin(x)=0
one way to do it is to write sin(x)cos(x)=0 <=> sin(x)=0 or cos(x)=0
<=> x=0, or x=pi/2 or x=pi or x=3pi/2 or x=2pi
A simplier way to write the solutions is just x=0[pi/2]
This means x=kpi/2, k being an integer and therefore you have ALL the solutions in one concise form.
However, I have to admit that this notation is more used in arithmetics problems, and that using them in trigonometry is just "laziness" in a way.
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