Применение основных тригонометрических формул для решения уравнений. Задача 44

ОТВЕТ: \(\dfrac{\pi }{2} + \pi k;\;\,\,\; \pm \dfrac{\pi }{3} + \pi k;\;\,\,\;k \in Z.\)

\(\cos 4x + 6{\cos ^2}x = 1\)

Воспользуемся формулой косинуса двойного угла:  \(\cos 2\alpha  = 2{\cos ^2}\alpha -1\)  и формулой понижения степени:  \({\cos ^2}\alpha  = \dfrac{{1 + \cos 2\alpha }}{2}.\)

\(\cos 4x + 6{\cos ^2}x = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,2{\cos ^2}2x-1 + 6 \cdot \dfrac{{1 + \cos 2x}}{2} = 1\,\,\,\,\,\,\, \Leftrightarrow \)

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

Пусть \(\cos 2x = t,\,\,\,\,\,\,t\, \in \,\left[ {-1;1} \right]\). Тогда:

\(2{t^2} + 3t + 1 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = -1,\,\,}\\{t = -\dfrac{1}{2}.}\end{array}} \right.\)

Вернёмся к прежней переменной:

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

Ответ: \(\dfrac{\pi }{2} + \pi k;\;\,\,\; \pm \dfrac{\pi }{3} + \pi k;\;\,\,\;k \in Z.\)