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The Wave Function: Solving Equations - HSN forum

# The Wave Function: Solving Equations

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Posted 27 April 2006 - 06:19 PM

root3cosx + sinx = root2

therefore 2cos(x - /6) = root2

cos(x - /6) = root2/2 = 1/root2

(x - /6) = /4 or 7/4

My question is, where does the 7/4 bit come from? I understand where the /4 comes from, but not 7/4. Can somebody please explain?

Thank-you.

### #2Mr H

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Posted 27 April 2006 - 07:15 PM

From the CAST rule.

45 is the acute angle.

Use the CAST diagram to find the second solution for positive cos.

So 2nd solution (in degrees for easier illustration) is 360 - 45 = 315 .

Which is 7 x 45 , in other words 7 /4.

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Posted 27 April 2006 - 08:47 PM

cosx - root3sinx = 1

k = 2
alpha = 300

So 2cos(x - 300) = 1

-> cos(x - 300) = 0.5
-> x - 300 = 60, 300?

(the final answers in this question are 0, 4/3 and 2 :s)

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Posted 27 April 2006 - 09:12 PM

QUOTE(Betinho @ Apr 27 2006, 09:47 PM)

cosx - root3sinx = 1

k = 2
alpha = 300

So 2cos(x - 300) = 1

-> cos(x - 300) = 0.5
-> x - 300 = 60, 300?

I hate these type, and I've never been told the proper way to solve them so I do it with a bit of common sense.

Say your looking for values of x that satisfy this equation with the range:

Now we have:

Now if we took the normal values for the inverse cosine of 0.5 we would get the angle of 60o, and we would say that within the range the values are 60o and 300o
However, we are adding 300o onto the angle, and if we add 300o onto 300o then we get a value outwith our range.
However, adding 300o onto 60o gives us a value still within the range so it's fine to use this
There is another value for x though, and if you think of this as coming from the cosine graph before zero (because it repeats, just as there is a repeat of the graph after 360 there is one before zero). So if we took -60 (i.e. the equivalent of the quadrant four angle in the graph before zero) as one of our values for x, added 300 onto this and got 240 this would give us a value still within our range. Also, if we took the quadrant one angle from our graph before zero, which is -300 (i.e. -360 + 60), and add 300 onto this we get zero which is still in our range.
So from this we get the results:
x = 0, 240, 360

That's how I do it anyway, not sure about how good a method it is
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### #5Mr H

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Posted 27 April 2006 - 09:16 PM

Too slow at typing!!

Following from your last line Betinho;

x = 360 , 600

As ad adsurdum states, 600 is outwith the usual 0 < x < 360 domain.

The period/wavelength is 360 as there is only a single 'x' therefore every 360 there will be another solution.

The 360 solution above is OK, there will also be a solution 360 before it, i.e. 0 , and also 360 before 600 i.e. 240 .

So the solutions are (in degrees) 0 , 240 , 360

Or 0, 4 /3, 2 in radians.

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### #6Mr H

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Posted 27 April 2006 - 09:27 PM

QUOTE(ad absurdum @ Apr 27 2006, 10:12 PM)
I hate these type, and I've never been told the proper way to solve them so I do it with a bit of common sense.

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Posted 27 April 2006 - 09:27 PM

Thanks guys!

It makes sense now. It was beginning to annoy me - I've been staring at that question for days.

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Posted 27 April 2006 - 09:49 PM

QUOTE(Mr H @ Apr 27 2006, 10:27 PM)

QUOTE(ad absurdum @ Apr 27 2006, 10:12 PM)
I hate these type, and I've never been told the proper way to solve them so I do it with a bit of common sense.