Однородные тригонометрические уравнения. Задача 19

Задача 19. Решите уравнение: \({\sin ^2}4x = 3{\cos ^2}4x\)

Ответ

ОТВЕТ: \( \pm \dfrac{\pi }{{12}} + \dfrac{{\pi k}}{4};\,\,\,\,\,\,\,k \in Z.\)

Решение

\({\sin ^2}4x = 3{\cos ^2}4x.\)

Однородное тригонометрическое уравнение второй степени:

\({\sin ^2}4x = 3{\cos ^2}4x\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\rm{t}}{{\rm{g}}^2}4x = 3\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{\rm{tg}}4x = \sqrt 3 ,\,\,\,}\\{{\rm{tg}}4x = -\sqrt 3 }\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{4x = \dfrac{\pi }{3} + \pi k,\,\,}\\{4x = -\dfrac{\pi }{3} + \pi k}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \dfrac{\pi }{{12}} + \dfrac{{\pi k}}{4},}\\{x = -\dfrac{\pi }{{12}} + \dfrac{{\pi k}}{4},}\end{array}\,} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = \pm \dfrac{\pi }{{12}} + \dfrac{{\pi k}}{4},\,\,\,\,\,\,\,\,\,k\, \in \,Z.\)

Ответ: \( \pm \dfrac{\pi }{{12}} + \dfrac{{\pi k}}{4};\,\,\,\,\,\,\,k \in Z.\)