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Hi guys!,

 

determine the general equation of \(Sin(x-30)=Cos 2x\)

 

My way is this:

 

\(sin(x-30)=sin(90-2x) \)

\(x-30=90-2x\)

\(3x=120\)

\(x=40\)

 

Obviously this is wrong as this is not an equation....The answer given is:

 

\(sin(x-30)=sin(90-2x)\)

\(x-30=90-2x + 360k\)

\(3x=120 + 360k\)

\(x=40+120k\)

 

please explain the addition to me?..thank you kindly..

 
juriemagic  Nov 8, 2018
edited by juriemagic  Nov 8, 2018
 #1
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The answer given is wrong.

 

In order to correctly find the solution the trig functions must all be expanded out and combined.

 

This results in a 4th degree polynomial in either sin(x) or cos(x) depending on how you expand things.

 

I was working on writing up the LaTex for the solution when the editor hiccupped on me and I lost it all.

 

See if you can reproduce it.  It's not difficult.  A couple of trig identities and some algebra.

 
Rom  Nov 8, 2018
 #2
avatar+278 
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Thank you Rom,

 

I do appreciate your time...

 
juriemagic  Nov 8, 2018
 #3
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This is my understanding of this:

 

Since these are co-functions, the angles add up to 90 degrees:

x - 30 + 2x = 90

3x =90 + 30

3x = 120 

x = 120 / 3 

x = 40 degrees.

 
Guest Nov 8, 2018
 #5
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Thank you very much guest,

 

i do apreciate your input...

 
juriemagic  Nov 12, 2018
 #4
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Hi Guest and Jurimagic,

 

Here are the graphs of y=LHS   and y=RHS

you can see that the LHS has a wavelength of 360 and the RHS has a wavelength of 180.

You should know that just by looking at the formulas.

 

So any solution k that you find between 0 and 360 will be repeated k+360n   where n is an integer.

From the graph you can see that there will be 4 solutions between x=0 and x=360.

You have only found the smallest one.

 

Hi Rom, you say that this is 4th degree polynomial is not hard to determine but I am having trouble with the trig. Maybe I am suffering brain freeze but it is not falling out for me.  Maybe you can help some more. Please  laugh

 

 
Melody  Nov 11, 2018
edited by Melody  Nov 11, 2018
 #6
avatar+278 
+1

Hello Melody,

 

gosh, you really have walked the extra mile!!!!..many many thanks!!!...I soooo wish I had just some of your knowledge!!!..once again, thank you..

 
juriemagic  Nov 12, 2018
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Ok... I guess I'm being really stupid or something.

Everything below is in degrees.

 

\(\sin(x) = \cos(x -90)\\ \sin(x-30) = \cos(x-120)\\ \text{we also have }\cos(x) = \cos(-x) \\ \text{and }\cos(2x + 360k)=\cos(2x)\)

as we move forward from x=0 the first place the equation is satisfied is at

 

\(\cos(2x) = \cos(120-x)\\ 3x=120\\ x=40\)

 

the second place the equation is satisfied is at

 

\(\cos(2x-360)=\cos(120-x)\\ 2x-360 = 120-x\\ 3x = 480 \\ x=160\)

 

the third place the equation is satisfied is at

 

\(\cos(2x-360) = \cos(x-120)\\ 2x-360=x-120\\ x=240\)

 

and the final place less than 2pi is at

 

\(\cos(720-2x) = \cos(x-120)\\ 720-2x=x-120\\ 840=3x\\ x = 280\)

 

So the solutions, as Melody showed graphically, are 

x = 40, 160, 240, 280 degrees.

 
Rom  Nov 12, 2018
edited by Rom  Nov 12, 2018
edited by Rom  Nov 12, 2018
edited by Rom  Nov 12, 2018
edited by Rom  Nov 12, 2018
edited by Rom  Nov 12, 2018
 #9
avatar+278 
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Rom,

 

thank you for coming back to an "old" post...I honestly do appreciate your time with this. Thanx again for the explanation..I wish you a great day!!!

 
juriemagic  Nov 13, 2018

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