# Thread: Finding the general solution (sin3x = sin2x)

1. ## Finding the general solution (sin3x = sin2x)

For some reason my answer is completely off the asnwers in the back of my book.

My method:

sin3x = sin2x
sin3x - sin2x = 0
but sin(A+B) - sin(A-B) = 2sinBcosA
A + B = 3x; -A + B = -2x
2A = 5x ; A = 5x/2
2B = x; B = x/2
Therefore sin(3x) - sin(2x) = 2sin(x/2)cos(5x/2) = 0
i.e.:

sin(x/2) = 0
x/2 = n(pi); x = 2n(pi) (n = 0, +/-1, +/-2,....)

OR cos(5x/2) = 0
(+/-)5x/2 = n(pi)/2
i.e. 5x/2 = n(pi/2) OR 5x/2 = -n(pi/2)
x = n(pi)/5 or x = -n(pi)/5
therefore x = +/- (n)(pi)/5
but n already includes the +/-;
therefore x = npi/5

Therefore, x = n(pi)/5 or x = 2npi
but n(pi)/5 includes 2n(pi)
therefore general solution: x = n(pi)/5

x = 2n(pi) or x = 4n/5 +/- pi/5

Any clue on what i'm doing wrong?

2. ## Re: Finding the general solution (sin3x = sin2x)

Originally Posted by namelyanonymous
For some reason my answer is completely off the asnwers in the back of my book.

My method:

sin3x = sin2x
sin3x - sin2x = 0
but sin(A+B) - sin(A-B) = 2sinBcosA
A + B = 3x; -A + B = -2x
2A = 5x ; A = 5x/2
2B = x; B = x/2
Therefore sin(3x) - sin(2x) = 2sin(x/2)cos(5x/2) = 0
i.e.:

sin(x/2) = 0
x/2 = n(pi); x = 2n(pi) (n = 0, +/-1, +/-2,....)

OR cos(5x/2) = 0
(+/-)5x/2 = n(pi)/2
i.e. 5x/2 = n(pi/2) OR 5x/2 = -n(pi/2)
x = n(pi)/5 or x = -n(pi)/5
therefore x = +/- (n)(pi)/5
but n already includes the +/-;
therefore x = npi/5

Therefore, x = n(pi)/5 or x = 2npi
but n(pi)/5 includes 2n(pi)
therefore general solution: x = n(pi)/5

x = 2n(pi) or x = 4n/5 +/- pi/5

Any clue on what i'm doing wrong?
I wouldn't be complicating things by using your method.
Just say 3x=2x+2n.pi, or 3x=(pi-2x) + 2n.pi then solve each for x.

You are saying the solution to cosx=0 is EVERY multiple of pi/2, instead of just the odd multiples.

3. ## Re: Finding the general solution (sin3x = sin2x)

$The general solution for the sine function is:$
$\sin x = \sin \alpha \Rightarrow x = \pi n + (-1)^n\alpha$.

4. ## Re: Finding the general solution (sin3x = sin2x)

Originally Posted by asianese
$The general solution for the sine function is:$
$\sin x = \sin \alpha \Rightarrow x = \pi n + (-1)^n\alpha$.
Which IS n.pi when alpha=0

BTW, I wish teachers would not teach those monstrosities. Its so much easier to just add 2n.pi to each solution at the point where the sin/cos/tan is removed.

5. ## Re: Finding the general solution (sin3x = sin2x)

Originally Posted by braintic
I wouldn't be complicating things by using your method.
Just say 3x=2x+2n.pi, or 3x=(pi-2x) + 2n.pi then solve each for x.

You are saying the solution to cosx=0 is EVERY multiple of pi/2, instead of just the odd multiples.
Gah I feel so stupid now for missing both of those points. Thanks a lot!

Edit: I'm still getting the wrong second answer for some reason ((pi)/5 + 2n(pi)/5) :/

6. ## Re: Finding the general solution (sin3x = sin2x)

Originally Posted by asianese
$The general solution for the sine function is:$
$\sin x = \sin \alpha \Rightarrow x = \pi n + (-1)^n\alpha$.
I have really horrible memory so I just avoid even trying to memorise stuff because I know I'll end up forgetting it under pressure :/.
Out of curiosity, would this really get you a full mark for questions a tad harder than this in the HSC?

7. ## Re: Finding the general solution (sin3x = sin2x)

Originally Posted by braintic
Which IS n.pi when alpha=0

BTW, I wish teachers would not teach those monstrosities. Its so much easier to just add 2n.pi to each solution at the point where the sin/cos/tan is removed.
I don't see why this is a 'monstrosity'. If a formula's derivation is taught and understood properly by students, it is not a sin to quote a formula and use it.

There are a myriad of ways of thinking about mathematics, formulae and methods, and it'd be fair to introduce students to as many as there are possible. In the case of general solutions for trig, one can 'draw the picture and tell the story', as my wonderful MX1 teacher would tell us - drawing a picture of the quadrants, figuring out base angles and then generalising to n integers. The 'monstrosity' is another interpretation which arises from the fact that the sine function is odd (-1)^n, and this fact is easily seen when drawing a horizontal line through a sine curve and observing in what multiples you skip around to touch the sine curve again. This leads to the more often quoted monstrosity ver2. of $\sin \theta = k \Rightarrow \theta = \pi n +(-1)^n \sin^{-1}k.$ which is in the syllabus (not trying to imply the syllabus is the best thing ever in history and it should be our bible of mathematics, but is what the students are tested on)

8. ## Re: Finding the general solution (sin3x = sin2x)

Originally Posted by namelyanonymous
I have really horrible memory so I just avoid even trying to memorise stuff because I know I'll end up forgetting it under pressure :/.
Out of curiosity, would this really get you a full mark for questions a tad harder than this in the HSC?
Depends on the mark scheme. If its 1-2 marks, using the formula would get full marks. It is in the syllabus, after all (Quoted in the form above)

9. ## Re: Finding the general solution (sin3x = sin2x)

Originally Posted by asianese
I don't see why this is a 'monstrosity'. If a formula's derivation is taught and understood properly by students, it is not a sin to quote a formula and use it.

There are a myriad of ways of thinking about mathematics, formulae and methods, and it'd be fair to introduce students to as many as there are possible. In the case of general solutions for trig, one can 'draw the picture and tell the story', as my wonderful MX1 teacher would tell us - drawing a picture of the quadrants, figuring out base angles and then generalising to n integers. The 'monstrosity' is another interpretation which arises from the fact that the sine function is odd (-1)^n, and this fact is easily seen when drawing a horizontal line through a sine curve and observing in what multiples you skip around to touch the sine curve again. This leads to the more often quoted monstrosity ver2. of $\sin \theta = k \Rightarrow \theta = \pi n +(-1)^n \sin^{-1}k.$
I didn't say I had a problem with people using it if they can learn it properly.
But after years of marking, I can tell you that the students who use these more complicated formulas on average make more mistakes.
My problem is with teachers who don't recognise this. They should at the very least introduce their students to the easiest method first before explaining the more complicated formulae.
In fact, some teachers refuse to even acknowledge that simply adding 2n.pi is a valid way or writing the general solution.

10. ## Re: Finding the general solution (sin3x = sin2x)

Originally Posted by braintic
I didn't say I had a problem with people using it if they can learn it properly.
But after years of marking, I can tell you that the students who use these more complicated formulas on average make more mistakes.
My problem is with teachers who don't recognise this. They should at the very least introduce their students to the easiest method first before explaining the more complicated formulae.
In fact, some teachers refuse to even acknowledge that simply adding 2n.pi is a valid way or writing the general solution.
I agree with this. After seeing the trouble/delusion/ohno moments of SIMPSONS rule (this is the biggest pain) of students, such formulae often are used incorrectly or are poorly understood. Some SWEAR by the (h/3) method, 2 of the evens, 4 of the odds, chop the rest off (i don't even know it), whilst others talk about simplifying life and use (b-a)/6.

The thing is these are all the same!! (as with this little general solution thing)

Back on topic: @OP: even if your answer looks different to the solution it might not necessarily be wrong (I'm talking about general solutions). Try checking your answer by comparing the two - sub in some n values and see if they work in the original equation sin2x=sin3x.

11. ## Re: Finding the general solution (sin3x = sin2x)

Originally Posted by braintic
Which IS n.pi when alpha=0

BTW, I wish teachers would not teach those monstrosities. Its so much easier to just add 2n.pi to each solution at the point where the sin/cos/tan is removed.
This.

except maybe in the case of the tan, we would add n*pi.

I very much dislike this formula and always teach the x+2kpi and pi-x +2kpi method.

12. ## Re: Finding the general solution (sin3x = sin2x)

To get the general solution, use the sin(a + b) formular with sin(3x) = sin(2x + x), Then manipulate the terms on both sides until you get the golden ratio forumula:
x^2 - x - 1 = 0, where x = 2 cosx.
Fascinating result: cosx = 1/2 (1 + sqrt(5))/2 and cosx = 1/2 (1 - sqrt(5))/2)
Another siting of the golden ratio.

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