Задача 58. Решите уравнение: \({\cos ^2}\dfrac{x}{2}-2{\cos ^2}\dfrac{{5x}}{2}-{\sin ^2}\dfrac{x}{2} + 1 = 0\)
ОТВЕТ: \(\dfrac{{\pi k}}{3};\;\,\,\;\dfrac{{\pi k}}{2};\,\,\,\,\,\,k \in Z.\)
\({\cos ^2}\dfrac{x}{2}-2{\cos ^2}\dfrac{{5x}}{2}-{\sin ^2}\dfrac{x}{2} + 1 = 0\) Воспользуемся формулами косинуса двойного угла: \(\cos 2\alpha = {\cos ^2}\alpha -{\sin ^2}\alpha ;\,\,\,\,\,\,\,\,\,\,\,\,\cos 2\alpha = 2{\cos ^2}\alpha -1.\) \({\cos ^2}\dfrac{x}{2}-2{\cos ^2}\dfrac{{5x}}{2}-{\sin ^2}\dfrac{x}{2} + 1 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\cos ^2}\dfrac{x}{2}-{\sin ^2}\dfrac{x}{2}-\left( {2{{\cos }^2}\dfrac{{5x}}{2}-1} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,\cos x-\cos 5x = 0.\) Воспользуемся формулой разности косинусов: \(\cos \alpha -\cos \beta = -2\sin \dfrac{{\alpha + \beta }}{2}\sin \dfrac{{\alpha -\beta }}{2}.\) \(\cos x-\cos 5x = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,-2\sin 3x\sin \left( {-2x} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sin 3x = 0,}\\{\sin 2x = 0\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{3x = \pi k,}\\{2x = \pi k\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \dfrac{{\pi k}}{3},}\\{x = \dfrac{{\pi k}}{2},}\end{array}} \right.\,\,\,\,\,\,\,\,\,\,\,\,k\, \in \,Z.\) Ответ: \(\dfrac{{\pi k}}{3};\;\,\,\;\dfrac{{\pi k}}{2};\,\,\,\,\,\,k \in Z.\)