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.