Pentru rapiditate, notam a+b=x
Avem [tex]sin(x+\dfrac{\pi}{6})=sinx\ cos \dfrac{\pi}{6}+cosx\ sin\dfrac{\pi}{6}\Rightarrow \dfrac{\sqrt3}{2}sinx+\dfrac12cosx=-\dfrac12[/tex]
Analog:
[tex]sin(x-\dfrac{\pi}{6})=sinx\ cos \dfrac{\pi}{6}-cosx\ sin\dfrac{\pi}{6}\Rightarrow \dfrac{\sqrt3}{2}sinx-\dfrac12cosx=\dfrac12[/tex]
Adunam cele doua relatii si obtinem:
[tex]\sqrt3\ sinx=0\Rightarrow sinx=0\Rightarrow x=a+b=k\pi;\ k\in\mathbb Z[/tex]
