Задача 11. Решите уравнение: \(\cos \,\left( {2\sin x + \left( {1 + \sqrt 3 } \right)\,\cos x\,} \right) = \sin \,\left( {\,\left( {1-\sqrt 3 } \right)\,\cos x\,} \right)\,\)
ОТВЕТ: \({\left( {-1} \right)^k}\arcsin \left( {\dfrac{{\pi \,\sqrt 2 }}{8}} \right)-\dfrac{\pi }{4} + \pi \,k;\,\,\,\,\,\,{\left( {-1} \right)^{k + 1}}\arcsin \dfrac{\pi }{8}-\dfrac{\pi }{3} + \pi \,k;\;\,\,\,\;k \in Z.\)
\(\cos \left( {2\sin x + \left( {1 + \sqrt 3 } \right)\cos x} \right) = \sin \left( {\left( {1-\sqrt 3 } \right)\cos x} \right)\,\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,\sin \left( {\dfrac{\pi }{2} + 2\sin x + \left( {1 + \sqrt 3 } \right)\cos x} \right) + \sin \left( {\left( {\sqrt 3 -1} \right)\cos x} \right) = 0.\) Воспользуемся формулой суммы синусов: \(\sin \alpha + \sin \beta = 2\sin \dfrac{{\alpha + \beta }}{2}\cos \dfrac{{\alpha -\beta }}{2}.\) \(2\sin \dfrac{{\dfrac{\pi }{2} + 2\sin x + \left( {1 + \sqrt 3 } \right)\cos x + \left( {\sqrt 3 -1} \right)\cos x}}{2}\cos \dfrac{{\dfrac{\pi }{2} + 2\sin x + \left( {1 + \sqrt 3 } \right)\cos x-\left( {\sqrt 3 -1} \right)\cos x}}{2} = 0\,\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sin \left( {\dfrac{\pi }{4} + \sin x + \sqrt 3 \cos x} \right) = 0,}\\{\cos \left( {\dfrac{\pi }{4} + \sin x + \cos x} \right) = 0\,\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\dfrac{\pi }{4} + \sin x + \sqrt 3 \cos x = \pi k,\,}\\{\dfrac{\pi }{4} + \sin x + \cos x = \dfrac{\pi }{2} + \pi k,}\end{array}} \right.} \right.\,\,\,\,\,\,\,k \in Z.\) Рассмотрим первое уравнение последней совокупности: \(\sin x + \sqrt 3 \cos x = -\dfrac{\pi }{4} + \pi k,\,\,\,k\, \in \,Z.\) Так как \(\sin x + \sqrt 3 \cos x\) принимает значение \(\left[ {-2;2} \right]\), то это уравнение будет иметь решение только при \(k = 0.\) Следовательно: \(\sin x + \sqrt 3 \cos x = -\dfrac{\pi }{4}\left| {:2} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{1}{2}\sin x + \dfrac{{\sqrt 3 }}{2}\cos x = -\dfrac{\pi }{8}\,\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,\cos \dfrac{\pi }{3}\sin x + \sin \dfrac{\pi }{3}\cos x = -\dfrac{\pi }{8}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\sin \left( {x + \dfrac{\pi }{3}} \right) = -\dfrac{\pi }{8}\,\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,x + \dfrac{\pi }{3} = {\left( {-1} \right)^{k + 1}}\arcsin \dfrac{\pi }{8} + \pi k\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = -\dfrac{\pi }{3} + {\left( {-1} \right)^{k + 1}}\arcsin \dfrac{\pi }{8} + \pi k,\,\,\,k\, \in \,Z.\) Рассмотрим второе уравнение: \(\sin x + \cos x = \dfrac{\pi }{4} + \pi k,\,\,\,k\, \in \,Z.\) Так как \(\sin x + \cos x\) принимает значение \(\left[ {-\sqrt 2 ;\sqrt 2 } \right]\), то это уравнение будет иметь решение только при \(k = 0.\) Следовательно: \(\sin x + \cos x = \dfrac{\pi }{4}\left| {:\sqrt 2 } \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{\sqrt 2 }}{2}\sin x + \dfrac{{\sqrt 2 }}{2}\cos x = \dfrac{\pi }{{4\sqrt 2 }}\,\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,\cos \dfrac{\pi }{4}\sin x + \dfrac{\pi }{4}\cos x = \dfrac{{\pi \sqrt 2 }}{8}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\sin \left( {x + \dfrac{\pi }{4}} \right) = \dfrac{{\pi \sqrt 2 }}{8}\,\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,x + \dfrac{\pi }{4} = {\left( {-1} \right)^k}\arcsin \dfrac{{\pi \sqrt 2 }}{8} + \pi k\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,x = -\dfrac{\pi }{4} + {\left( {-1} \right)^k}\arcsin \dfrac{{\pi \sqrt 2 }}{8} + \pi k,\,\,\,k\, \in \,Z.\) Ответ: \(-\dfrac{\pi }{3} + {\left( {-1} \right)^{k + 1}}\arcsin \dfrac{\pi }{8} + \pi k;\,\,\,\,-\dfrac{\pi }{4} + {\left( {-1} \right)^k}\arcsin \dfrac{{\pi \sqrt 2 }}{8} + \pi k,\,\,\,\,k\, \in \,Z.\)