Solving $\sqrt{3} \sin(\phi-\pi/6)=\sin(\phi)$

The question is to solve: $$\sqrt{3} \sin \left ( \phi - \frac \pi 6 \right )= \sin \phi$$

I tried to turn it into $a \sin \phi-b \cos\phi= \sin\phi$,

Then I got $\frac 32 \sin\phi - \frac 3 4 \cos \phi= \sin\phi$,

Therefore, $\frac 12 \sin\phi- \frac 3 4 \cos\phi=0$.

Then I turned it back into $R\sin(\phi-a)=0$,

$\implies (√13/4)\sin(ø-0.983)=0$

So I think the solution should be $\phi=n\pi+0.983$

Is that correct? The answer on my book is $n\pi+\frac \pi 3$.(I made a mistake here, It should be π/3.)

Update: The reason why I got the wrong answer is because I forgot to √ that 4/3.

Thanks for pointing that mistake out :D.

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The coefficient of $-3/4$ in front of $\cos\phi$ is incorrect. It should be $-\sqrt{3}/2$. –  anon Jun 27 '12 at 7:31

It may be directly written that: $$\sqrt3 \cos\phi= \sin \phi \implies \tan\phi = \sqrt{3} \implies \phi = n \pi + \dfrac{\pi}3 \text{ where }n\in \mathbb{Z}$$
$$\sqrt{3} \sin(\phi - \pi/6) = \sin(\phi) \implies \sqrt{3} \left( \sin(\phi) \cos(\pi/6) - \cos(\phi) \sin(\pi/6)\right) = \sin(\phi)$$ This gives us $$\sqrt{3}\sin(\phi) \times \sqrt{3}/2 - \sqrt{3} \cos(\phi)\times1/2 = \sin(\phi) \implies 3/2 \sin(\phi) - \sin(\phi) = \sqrt{3}/2 \cos(\phi)\\ \implies \dfrac{\sin(\phi)}2 = \dfrac{\sqrt{3}}2 \cos(\phi) \implies \tan(\phi) = \sqrt{3} \implies \phi = n \pi + \dfrac{\pi}3 \text{ where }n\in \mathbb{Z}$$ The mistake you have made is that your $b$ should be $\sqrt{3}/2$ and not $3/4$ as you have written. Also, the answer given in your book answer is incorrect, assuming you have stated the problem correctly.