\({\sin ^2}\dfrac{x}{2}-{\cos ^2}\dfrac{x}{2} = \dfrac{{\sqrt 3 }}{2}\)
Воспользуемся формулой косинуса двойного угла:
\(\cos 2\alpha = {\cos ^2}\alpha -{\sin ^2}\alpha .\)
\({\sin ^2}\dfrac{x}{2}-{\cos ^2}\dfrac{x}{2} = \dfrac{{\sqrt 3 }}{2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\cos ^2}\dfrac{x}{2}-{\sin ^2}\frac{x}{2} = -\dfrac{{\sqrt 3 }}{2}\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\cos \left( {2 \cdot \dfrac{x}{2}} \right) = -\dfrac{{\sqrt 3 }}{2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\cos x = -\dfrac{{\sqrt 3 }}{2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = \pm \dfrac{{5\pi }}{6} + 2\pi k,\,\,\,\,\,\,k\, \in \,Z.\)
Ответ: \( \pm \dfrac{{5\pi }}{6} + 2\pi k;\quad k \in Z.\)