Применение основных тригонометрических формул для решения уравнений. Задача 43

ОТВЕТ: \(\pi k;\;\,\,\; \pm \dfrac{\pi }{6} + \pi k;\;\,\,\;k \in Z.\)

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

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

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

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

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:

\(\left[ {\begin{array}{*{20}{c}}{\cos 2x = 1,}\\{\cos 2x = \dfrac{1}{2}}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{2x = 2\pi k,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{2x =  \pm \dfrac{\pi }{3} + 2\pi k}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \pi k,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x =  \pm \dfrac{\pi }{6} + \pi k,}\end{array}} \right.\,\,\,\,\,\,\,\,\,\,\,k\, \in \,Z.\)

Ответ: \(\pi k;\;\,\,\; \pm \dfrac{\pi }{6} + \pi k;\;\,\,\;k \in Z.\)