Sin 3x + sin 6x + sin 9x = 0 - Solve, giving values from 0 -360. (Half completed need help with second part please!)

Question

jadedry · Answer

First off i converted it to:

$$2\sin6x \cos3x +\sin 6x = 0$$

Therefore, Sin 6x or Cos 3x = -0.5 .

for Sin 6X, x = $$\frac{ n \pi }{ 6 }$$ = 0,30,60,90,120....360

For Cos 3X to make the entire equation 0 = 3x = 120, x = 40

But from this point forward i'm a bit confused as to how to proceed? I thought I would have values of 40, 120, 200. But apparently not?

Thanks in advance!

sirm3d · Answer

$$2\sin6x\cos3x+\sin6x=0$$
$$\sin6x\left\{2\cos3x+1\right\}=0$$
$$\sin6x=0,2\cos3x+1=0$$
solve the two equations separately

sirm3d · Answer

also, when solving $\sin6x=0$ for $0\le x<360$, you should consider the range of values $0\le 6x <6\times360$

jadedry · Answer

I had those equations at one point, I'm still confused about cos though, if cos 3x + 1 = 0 then x = 60 which is wrong according to my book? Maybe I'm misunderstanding you?

sirm3d · Answer

$$2\cos3x+1=0$$ $$\cos3x=-\frac{1}{2}$$ $$3x=120,240,480,600,840,960$$
therefore $$x=40,80,160,200,280,320$$

jadedry · Answer

Ah! I understand. I was dividing it before I came up with a list of numbers. Thank you. c:

sirm3d · Answer

when $\sin6x=0$ then $$6x=0,180,360,540, 720,900,1080,1260,1440,1620,1800,1980$$ or 
$$x=0,300,60,90, 120,150,180,210,240,270,300,330$$

sirm3d · Answer

you're welcome

jadedry · Answer

Alright, gotcha, thanks again! c: