Vă rog, o cheie.. ce formula să folosesc?
Jos scrie "de arătat"

Aplici
sin(a-b) =sinacosb-sinbcosa
cos(a+b) =cosacosb-sinasinb
cospi/3=sinpi/6=1/2
Sinpi/3=cospi/6=rad3/2
Rezulta
1/2*sinx-rad3/2*cosx+rad3/2*cosx-1/2*sinx=0
Deci este vorba despre a demonstra o egalitate( de aici indicatia,, aratati"), nu de rezolv. o ecuatie!

[tex] \it sin\left(x-\dfrac{\pi}{3}\right) = sinxcos\dfrac{\pi}{3} -sin\dfrac{\pi}{3}cosx = \dfrac{1}{2} sinx - \dfrac{\sqrt3}{2} cosx
\\ \\ \\
cos\left(x+\dfrac{\pi}{6}\right) = cosxcos\dfrac{\pi}{6} -sinx sin\dfrac{\pi}{6}= \dfrac{\sqrt3}{2} cosx - \dfrac{1}{2} sinx
\\ \\ \\
sin\left(x-\dfrac{\pi}{3}\right) +cos\left(x+\dfrac{\pi}{6}\right) = \dfrac{1}{2} sinx - \dfrac{\sqrt3}{2} cosx +\dfrac{\sqrt3}{2} cosx - \dfrac{1}{2} sinx =0 [/tex]