Однородные тригонометрические уравнения. Задача 14

ОТВЕТ: \(\dfrac{\pi }{{10}} + \dfrac{{\pi k}}{5};\,\,\,\,\,\,\dfrac{1}{5}{\text{arctg}}\,\dfrac{1}{2} + \dfrac{{\pi k}}{5};\,\,\,\,\,k \in Z.\)

\({\cos ^2}5x-2\cos 5x\sin 5x = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\cos 5x\left( {\cos 5x-2\sin 5x} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\cos 5x = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\cos 5x-2\sin 5x = 0.}\end{array}} \right.\)

\(\cos 5x = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,5x = \dfrac{\pi }{2} + \pi k\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = \dfrac{\pi }{{10}} + \dfrac{{\pi k}}{5},\,\,\,\,\,\,\,\,\,k\, \in \,Z.\)

Уравнение \(\cos 5x-2\sin 5x = 0\) является однородным тригонометрическим уравнением первой степени:

\(\cos 5x-2\sin 5x = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,1-2{\rm{tg}}5x = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\rm{tg}}5x = \dfrac{1}{2}\,\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\,\,5x = {\rm{arctg}}\dfrac{1}{2} + \pi k\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = \dfrac{1}{5}{\rm{arctg}}\dfrac{1}{2} + \dfrac{{\pi k}}{5},\,\,\,\,\,\,\,\,\,k\, \in \,Z.\)

Ответ:  \(\dfrac{\pi }{{10}} + \dfrac{{\pi k}}{5};\,\,\,\,\,\,\dfrac{1}{5}{\rm{arctg}}\,\dfrac{1}{2} + \dfrac{{\pi k}}{5};\,\,\,\,\,k \in Z.\)