Уравнение вида a sin x + b cos x = c. Задача 9

ОТВЕТ: \(\dfrac{\pi }{6} + \dfrac{{2\pi k}}{3};\,\,\,\,\,\dfrac{{7\pi }}{{18}} + \dfrac{{2\pi k}}{3};\,\,\,\,\,k \in Z.\)

\(\sin 3x-\sqrt 3 \cos 3x = 1.\)

Уравнение вида:  \(a\sin x + b\cos x = c,\;\;\)где \(\;a = 1,\,\,\,\,b = -\sqrt 3 ,\;\;c = 1.\)

Разделим обе части исходного уравнения на  \(\sqrt {{a^2} + {b^2}} ,\;\) то есть на  \(\sqrt {{1^2} + {{\left( {-\sqrt 3 } \right)}^2}}  = 2.\)

\(\dfrac{1}{2}\sin 3x-\dfrac{{\sqrt 3 }}{2}\cos 3x = \dfrac{1}{2}\;\;\;\; \Leftrightarrow \;\;\;\;\cos \dfrac{\pi }{3}\sin 3x-\sin \dfrac{\pi }{3}\cos 3x = \dfrac{1}{2}\;\;\;\; \Leftrightarrow \)

\( \Leftrightarrow \;\;\;\;\sin \left( {3x-\dfrac{\pi }{3}} \right) = \dfrac{1}{2}\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{3x-\dfrac{\pi }{3} = \dfrac{\pi }{6} + 2\pi k,}\\{3x-\dfrac{\pi }{3} = \dfrac{{5\pi }}{6} + 2\pi k}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \dfrac{\pi }{6} + \dfrac{{2\pi k}}{3},\,\,\,}\\{x = \dfrac{{7\pi }}{{18}} + \dfrac{{2\pi k}}{3},}\end{array}} \right.\,\,\,\,\,k \in Z.\)

Ответ:  \(\dfrac{\pi }{6} + \dfrac{{2\pi k}}{3};\,\,\,\,\,\dfrac{{7\pi }}{{18}} + \dfrac{{2\pi k}}{3};\,\,\,\,\,k \in Z.\)