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

#1**+1 **

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

#3**+1 **

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

#4**0 **

**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

Melody
Nov 11, 2018

#6**+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

#7**0 **

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

#9**0 **

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