Trigonometry mixup!

We have $\sin \left (15\degree-\dfrac{x}{2}\right ) =\dfrac{2\cos 50\degree}{\sqrt 3\sin 80\degree}-\tan 10\degree=$

$\dfrac{\cos 50\degree}{\sin 60\degree\sin 80\degree}-\dfrac{\sin 10\degree}{\sin 80\degree}=$

$\dfrac{\cos 50\degree-\sin 60\degree\sin 10\degree}{\sin 60\degree\sin 80\degree}=$

$\dfrac{\cos 60\degree\cos 10\degree}{\sin 60\degree\cos 10\degree}=\dfrac{1}{\sqrt 3}$ .

So, $\sin (60\degree+x)=$

$\cos (30\degree-x)=1-2\sin^2 \left (15\degree-\dfrac{x}{2}\right ) =$

$1-2\times \dfrac{1}{3}=\boxed {\dfrac{1}{3}}$ .

$\begin{aligned} \sin\left(15\degree - \frac x2\right) & = \frac {2\cos 50\degree}{\sqrt 3\sin 80\degree} - \tan 10\degree \\ & = \frac {\sin 40\degree}{\frac {\sqrt 3}2\cos 10\degree} - \frac {\sin 10\degree}{\cos 10\degree} \\ & = \frac {\sin (30\degree+10\degree)- \frac {\sqrt 3}2\sin 10\degree}{\frac {\sqrt 3}2\cos 10\degree} \\ & = \frac {\frac 12 \cos 10\degree + \frac {\sqrt 3}2\sin 10\degree - \frac {\sqrt 3}2\sin 10\degree}{\frac {\sqrt 3}2\cos 10\degree} \\ & = \frac {\frac 12 \cos 10\degree}{\frac {\sqrt 3}2\cos 10\degree} = \frac 1{\sqrt 3} \end{aligned}$

Therefore, $\sin (60\degree + x) = \cos (30\degree - x) = 1 - 2\sin^2 \left(15\degree -\dfrac x2\right) = 1 - \dfrac 23 = \boxed{\dfrac 13}$ .