How to Solve a Challenging Trig Equation with Identities?

[kscplay]

Homework Statement

2sin(4x)-sin(2x)-(√3)cos(2x)=0

x is [0,2π]

Homework Equations

The Attempt at a Solution

Using trig identities:
8sinxcos³x-6sinxcosx-√3(2cos²x-1)

Please help me with this problem. I have no idea where to go from here. Thanks.

 

[kscplay]

I've further simplified it to:

4sin2x−tan2x−√3=0

Does this make it easier? Please help.

 

[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...

Using trig identities:
8sinxcos³x-6sinxcosx-√3(2cos²x-1)

This doesn't seem to be making things simpler.

4sin2x−tan2x−√3=0

Does this make it easier? Please help.

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!

 

[ehild]

sin(4x)=sin(2x+π/3)

When sin(α)=sin(β) either

α=β+2kπ

or

α=(π-β)+2kπ.

ehild

 

[micromass]

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.

 

[kscplay]

Thanks guys :)