# Challenging Trig Equation

1. Feb 15, 2012

### kscplay

1. The problem statement, all variables and given/known data
2sin(4x)-sin(2x)-(√3)cos(2x)=0

x is [0,2π]
2. Relevant equations

3. The attempt at a solution
Using trig identities:
8sinxcos3x-6sinxcosx-√3(2cos2x-1)

Last edited: Feb 15, 2012
2. Feb 15, 2012

### kscplay

I've further simplified it to:

4sin2x−tan2x−√3=0

Last edited: Feb 15, 2012
3. Feb 16, 2012

### Mentallic

I was going to give this a try but had to head off to work, but now that I'm looking at it I'm honestly stuck as well...

This doesn't seem to be making things simpler.

I got here also and couldn't do much with it.

I also converted $$\sin(2x)+\sqrt{3}\cos(2x)$$ into the $$R\sin(2x+\theta)$$ form and got $$2\sin(2x+\frac{\pi}{3})$$

so what needs to be solved now is

$$\sin(4x)=\sin(2x+\frac{\pi}{3})$$

But I'm kind of stuck here as well.

We'll get back to you!

4. Feb 16, 2012

### ehild

sin(4x)=sin(2x+π/3)

When sin(α)=sin(β) either

α=β+2kπ

or

α=(π-β)+2kπ.

ehild

Last edited: Feb 16, 2012
5. Feb 16, 2012

### micromass

Staff Emeritus
First, make everything in the angle 2x. This gives (as you already know)

$$4\sin(2x)\cos(2x)-\sin(2x)-(√3)\cos(2x)=0$$

Now the trick is to use Weierstrass substitution (which is a handy technique in integration but it works here too).

Let $t=\tan(x)$, then

$$\sin(2x)=\frac{2t}{1+t^2},~\cos(2x)=\frac{1-t^2}{1+t^2}$$

this will give a polynomial in t which should be easier to solve.

Not saying that this is the easiest way, but it gives a solution nonetheless.

6. Feb 17, 2012

### kscplay

Thanks guys :)