[tex]\sin x\cdot\cos x-\frac{\sqrt[]{3}}{2}\cos x=0[/tex][tex]\cos x(\sin x-\frac{\sqrt[]{3}}{2})=0[/tex][tex]\begin{gathered} \cos x=0 \\ x=\frac{\pi}{2},\frac{3\pi}{2} \end{gathered}[/tex][tex]\begin{gathered} \sin x-\frac{\sqrt[]{3}}{2}=0 \\ \sin x=\frac{\sqrt[]{3}}{2} \\ x=\frac{\pi}{3} \end{gathered}[/tex]
so the answer is pi/2, 3pi/2, and pi/3
