Применение основных тригонометрических формул для решения уравнений. Задача 55

ОТВЕТ: \(\dfrac{\pi }{8} + \dfrac{{\pi k}}{4};\;\,\,\;\dfrac{\pi }{4} + \dfrac{{\pi k}}{2};\,\,\,\,\,\,k \in Z.\)

\(1 + \cos 2x = 2{\sin ^2}3x\)

Воспользуемся формулой понижения степени:

\({\sin ^2}\alpha  = \dfrac{{1-\cos 2\alpha }}{2}.\)

\(1 + \cos 2x = 2{\sin ^2}3x\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,1 + \cos 2x = 2 \cdot \dfrac{{1-\cos 6x}}{2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\cos 6x + \cos 2x = 0.\)

Воспользуемся формулой суммы косинусов:

\(\cos \alpha  + \cos \beta  = 2\cos \dfrac{{\alpha  + \beta }}{2}\cos \dfrac{{\alpha -\beta }}{2}.\)

\(\cos 6x + \cos 2x = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,2\cos 4x\cos 2x = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\cos 4x = 0,}\\{\cos 2x = 0}\end{array}\,\,\,\,\,\,\, \Leftrightarrow } \right.\)

\( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{4x = \dfrac{\pi }{2} + \pi k,}\\{2x = \dfrac{\pi }{2} + \pi k\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \dfrac{\pi }{8} + \dfrac{{\pi k}}{4},}\\{x = \dfrac{\pi }{4} + \dfrac{{\pi k}}{2},}\end{array}} \right.\,\,\,\,\,\,\,\,\,\,\,\,k\, \in \,Z.\)

Ответ: \(\dfrac{\pi }{8} + \dfrac{{\pi k}}{4};\;\,\,\;\dfrac{\pi }{4} + \dfrac{{\pi k}}{2};\,\,\,\,\,\,k \in Z.\)