trigonometry - 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.

Answer

It may be directly written that:

$$\sqrt3 \cos\phi= \sin \phi \implies \tan\phi = \sqrt{3} \implies \phi = n \pi + \dfrac{\pi}3 \text{ where }n\in \mathbb{Z}$$

Q.E.D.