I know I'm supposed to do $\sin(3x) - \sin x = 0$ but beyond that I'm stuck.. I tried expanding $\sin(3x)$ but that didn't help.
- I want the value of $x$ in the interval $[0, 2\pi)$
I know I'm supposed to do $\sin(3x) - \sin x = 0$ but beyond that I'm stuck.. I tried expanding $\sin(3x)$ but that didn't help.
We have $\sin{3x} = 3\sin{x} - 4\sin^{3}{x}$ which says that we have to solve the equation $$3\sin{x} - 4\sin^{3}{x} - \sin{x}=0$$, that is $2 \sin{x} - 4\sin^{3}{x}=0$. Take $y = \sin{x}$ and so you have $$2y-4y^{3}=0 \Longrightarrow 2y(1-2y^{2})=0$$ and then see what happens. I hope this helps you out.
Or you can even try this $$\sin{3x} - \sin{x} = 2 \cos\Bigl(\frac{3x+x}{2}\Bigr) \cdot \sin\Bigl(\frac{3x-x}{2}\Bigr) = 2\cos{2x} \cdot \sin{x}$$
You could use the fact that $\sin x=\sin y$ if and only if either $x-y$ is an even integer times $\pi$ or $x+y$ is an odd integer times $\pi$.