auxillilary method (1 Viewer)

[maths lover]

im just doing my homework and i just cant seem to do this question it is doing my head in.
2sinx-2cosx=root6 for x between 0 and 360 thaxs

 

[Shadowdude]

WolframAlpha it.

As a side note - see if you can spot the mistake in this working (just to try your mathematical mind)

2sin(x)-2cos(x) = 6^(1/2)
2(sin(x)-cos(x)) = sqrt(6)
sin(x)-cos(x) = sqrt(6)/2
(sin(x)-cos(x))^2 = (sqrt(6)/2)^2
sin^2(x) - 2sin(x)cos(x) + cos^2(x) = 6/4 (square both sides)
1-2sin(x)cos(x) = 6/4 (as sin^2 + cos^2 = 1)
1-sin(2x) = 3/2 (as sin(2x) = 2sin(x)cos(x)
-sin(2x) = 1/2
sin(2x)= -1/2

and so on...

EDIT: Nevermind, I figured you'd might as well get the right answer instead of me playing games.

 

[maths lover]

2sin(x)-2cos(x) = 6^(1/2)
2(sin(x)-cos(x)) = sqrt(6)
sin(x)-cos(x) = sqrt(6)/2
(sin(x)-cos(x))^2 = (sqrt(6)/2)^2
sin^2(x) - 2sin(x)cos(x) + cos^2(x) = 6/4 (square both sides)
-2sin(x)cos(x) = 6/4 (as sin^2 + cos^2 = 1)
-sin(2x) = 6/4 (as sin(2x) = 2sin(x)cos(x)
sin(2x) = -6/4

no solution.

i really dont know but i think the secound last line is wrong.

 

[Shadowdude]

Yes, there's a disappearing one.

Anyway, the solution you should get is: -1/2 = sin(2x) and then 2x = arcsin(-1/2)

Solve from there.

 

[Drongoski]

If I'm not mistaken, 2 principal values of x are:

 

[kooliskool]

Erm, for the answer, it is only a partial answer out of , because when you square it, you created possible solutions for and , then you will have to substitute the answer back to the original equation to find out which one it is. So often, it's better to avoid squaring both side or dividing anything related to x when you are solving equations (unless you consider the cases).

So the best method to solve questions like this is usually solved by auxiliary method, as you said in your title, just this one is a bit special and you can solve without it (although it's longer using square). So here goes the auxiliary method:

Let:


Hence equating coefficients:

......1
......2