Find the intersection of $~\sin(2x)~$ and $~\sin(4x)~$

Question

Arcing, I obviously get $$4x=2x +2k\pi, ~~~~~~~~~~ 2x=2k\pi,~~~~~~~~~~~~~ x=k\pi$$ but my book also found other results and I don't understand the workings out: $$(4x=\pi-2x +2k\pi, ~~~~~\text{so that}~~~~~~~~~~~~ x=(\frac{\pi}{6})(1+2k))~?$$

Answer 1

The $\sin$ function has two simmetries:

• one is the $2\pi$ periodicity that you are using $\sin(x)=\sin(x+2\pi)$;
• the second is the fact that $\sin(x)=\sin(\pi-x)$.

If you exploit the second combined with the periodicity, you get exactly what is done in the proposed solution.

Answer 2

Assuming you want to solve for $x$ in $$\sin 4x -\sin 2x=0,$$ turn this first into $$2\cos{\left(\frac{4x+2x}{2}\right)} \sin{\left(\frac{4x-2x}{2}\right)}=0,$$ or $$\cos 3x\sin x=0,$$ which implies $\cos 3x=0$ or $\sin x=0.$ Thus, we have that $$3x=(2j+1)\fracπ2,$$ or $$x=πk,$$ where $j,\,k$ are arbitrary integers.

Answer 3

You maybe forgot that $$\sin(x)=\sin(\pi-x)$$

and

$$\sin(x)=\sin(y)\iff x-y=2k\pi\lor x+y=(2k+1)\pi.$$

---

You also have

$$\sin(4x)-\sin(2x)=2\cos(2x)\sin(2x)+\sin(2x)=(2\cos(2x)+1)\sin(2x)$$

and you cancel both factors separately.