Determine general equation

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.

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.

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  

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.