\(\cos 5x\cos 3x + \sin 5x\sin 3x = \dfrac{1}{2}\)
Воспользуемся формулой косинуса разности:
\(\cos \left( {\alpha -\beta } \right) = \cos \alpha \cos \beta + \sin \alpha \sin \beta .\)
\(\cos 5x\cos 3x + \sin 5x\sin 3x = \dfrac{1}{2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\cos \left( {5x-3x} \right) = \dfrac{1}{2}\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\cos 2x = \dfrac{1}{2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,2x = \pm \dfrac{\pi }{3} + 2\pi k\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = \pm \dfrac{\pi }{6} + \pi k,\,\,\,\,\,k\, \in \,Z.\)
Ответ: \(\, \pm \,\dfrac{\pi }{6} + \pi k;\quad \,k \in Z.\)