Тригонометрические уравнения повышенной сложности. Задача 9

ОТВЕТ: \(\dfrac{\pi }{4} + \pi \,k;\;\,\,\,\,\;k \in Z.\)

\({\sin ^4}x + {\cos ^4}x = \sin 2x-0,5\,\,\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\,\,{\left( {{{\sin }^2}x} \right)^2} + 2{\sin ^2}x{\cos ^2}x + {\left( {{{\cos }^2}x} \right)^2}-2{\sin ^2}x{\cos ^2}x = \sin 2x-\dfrac{1}{2}\,\,\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\,\,{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2}-\dfrac{1}{2}{\sin ^2}2x-\sin 2x + \dfrac{1}{2} = 0\,\,\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\,\,1-\dfrac{1}{2}{\sin ^2}2x-\sin 2x + \dfrac{1}{2} = 0\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\sin ^2}2x + 2\sin 2x-3 = 0\,\,\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sin 2x = 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\sin 2x = -3 \notin \left[ {-1;1} \right]}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\sin 2x = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,2x = \dfrac{\pi }{2} + 2\pi k\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,x = \dfrac{\pi }{4} + \pi k,\,\,\,k\, \in \,Z.\)

Ответ:  \(\dfrac{\pi }{4} + \pi k,\,\,\,k\, \in \,Z.\)