\(\sqrt 2 \,\sin \left( {\dfrac{\pi }{4} + \dfrac{x}{2}} \right)-\sin \dfrac{x}{2} = \dfrac{{\sqrt 2 }}{2}\)
Воспользуемся формулой синуса суммы:
\(\sin \left( {\alpha + \beta } \right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta .\)
\(\sqrt 2 \left( {\sin \dfrac{\pi }{4}\cos \dfrac{x}{2} + \cos \dfrac{\pi }{4}\sin \dfrac{x}{2}} \right)-\sin \dfrac{x}{2} = \dfrac{{\sqrt 2 }}{2}\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\sqrt 2 \cdot \dfrac{{\sqrt 2 }}{2}\cos \dfrac{x}{2} + \sqrt 2 \cdot \dfrac{{\sqrt 2 }}{2}\sin \dfrac{x}{2}-\sin \dfrac{x}{2} = \dfrac{{\sqrt 2 }}{2}\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\cos \dfrac{x}{2} + \sin \dfrac{x}{2}-\sin \dfrac{x}{2} = \dfrac{{\sqrt 2 }}{2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\cos \dfrac{x}{2} = \dfrac{{\sqrt 2 }}{2}\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\dfrac{x}{2} = \pm \dfrac{\pi }{4} + 2\pi k\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = \pm \dfrac{\pi }{2} + 4\pi k,\,\,\,\,\,\,\,k\, \in \,Z.\)
Ответ: \(\, \pm \,\dfrac{\pi }{2} + 4\pi k;\quad \,k \in Z.\)