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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 $[0, 2\pi)$
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up vote 7 down vote accepted

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 $$\sin{3x} - \sin{x} = 2 \cos\Bigl(\frac{3x+x}{2}\Bigr) \cdot \sin\Bigl(\frac{3x-x}{2}\Bigr) = 2\cos{2x} \cdot \sin{x}$$

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ow silly me i thought sin3x = 3sinx - 4cos^3(x) – andrei Sep 30 '10 at 19:08
Can i divide by y ? and get 2 - 4y^2 ? – andrei Sep 30 '10 at 19:10
@Andrei: You should think on your own, from now! – anonymous Sep 30 '10 at 19:26
Yes thank you! i try to think on my on from now sorry >.< – andrei Sep 30 '10 at 20:06
@andrei: regarding dividing by y, consider whether there any numbers that you cannot divide by. – Isaac Sep 30 '10 at 20:27

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 $\pi$.

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