\(\cos x\cos 2x-\sin x\sin 2x = -\dfrac{1}{2}\)
Воспользуемся формулой косинуса суммы:
\(\cos \left( {\alpha + \beta } \right) = \cos \alpha \cos \beta -\sin \alpha \sin \beta .\)
\(\cos x\cos 2x-\sin x\sin 2x = -\dfrac{1}{2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\cos \left( {x + 2x} \right) = -\dfrac{1}{2}\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\cos 3x = -\dfrac{1}{2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,3x = \pm \dfrac{{2\pi }}{3} + 2\pi k\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = \pm \dfrac{{2\pi }}{9} + \dfrac{{2\pi k}}{3},\,\,\,\,\,k\, \in \,Z.\)
Ответ: \(\, \pm \,\dfrac{{2\pi }}{9} + \dfrac{{2\pi k}}{3};\quad \,k \in Z.\)