Уравнение вида a sin x + b cos x = c. Задача 18

ОТВЕТ: \(\dfrac{1}{3}\arcsin \dfrac{2}{{\sqrt 5 }} + \dfrac{1}{3}\arcsin \dfrac{1}{{\sqrt 5 }} + \dfrac{{2\pi k}}{3};\,\,\,\,\,\,\,\dfrac{\pi }{3}-\dfrac{1}{3}\arcsin \dfrac{2}{{\sqrt 5 }} + \dfrac{1}{3}\arcsin \dfrac{1}{{\sqrt 5 }} + \dfrac{{2\pi k}}{3};\,\,\,\,\,k \in Z.\)

\(2\sin 3x-\cos 3x = 2.\)

Уравнение вида:  \(a\sin x + b\cos x = c,\;\;\)где \(\;a = 2,\,\,\,\,b = -1,\;\;c = 2.\)

Разделим обе части исходного уравнения на  \(\sqrt {{a^2} + {b^2}} ,\;\) то есть на  \(\sqrt {{2^2} + {{\left( {-1} \right)}^2}}  = \sqrt 5 .\)

\(\dfrac{2}{{\sqrt 5 }}\sin 3x-\dfrac{1}{{\sqrt 5 }}\cos 3x = \dfrac{2}{{\sqrt 5 }}.\)

Так как  \({\left( {\dfrac{2}{{\sqrt 5 }}} \right)^2} + {\left( {\dfrac{1}{{\sqrt 5 }}} \right)^2} = 1,\)  то пусть  \(\cos \varphi  = \dfrac{2}{{\sqrt 5 }},\,\,\,\sin \varphi  = \dfrac{1}{{\sqrt 5 }}.\)  Тогда уравнение примет вид:

\(\dfrac{2}{{\sqrt 5 }}\sin 3x-\dfrac{1}{{\sqrt 5 }}\cos 3x = \dfrac{2}{{\sqrt 5 }}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\cos \varphi \sin 3x-\sin \varphi \cos 3x = \dfrac{2}{{\sqrt 5 }}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\sin \left( {3x-\varphi } \right) = \dfrac{2}{{\sqrt 5 }}\,\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{3x-\varphi  = \arcsin \dfrac{2}{{\sqrt 5 }} + 2\pi k,\,\,\,\,\,\,}\\{3x-\varphi  = \pi -\arcsin \dfrac{2}{{\sqrt 5 }} + 2\pi k}\end{array}} \right.\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \dfrac{1}{3}\arcsin \dfrac{2}{{\sqrt 5 }} + \dfrac{\varphi }{3} + \dfrac{{2\pi k}}{3},\,\,\,\,\,\,\,\,\,}\\{x = \dfrac{\pi }{3}-\dfrac{1}{3}\arcsin \dfrac{2}{{\sqrt 5 }} + \dfrac{\varphi }{3} + \dfrac{{2\pi k}}{3},}\end{array}} \right.\,\,\,\,k \in Z.\)

Так как  \(\sin \varphi  = \dfrac{1}{{\sqrt 5 }},\)  то   \(\varphi  = \arcsin \dfrac{1}{{\sqrt 5 }}\)  и  

\(\left[ {\begin{array}{*{20}{c}}{x = \dfrac{1}{3}\arcsin \dfrac{2}{{\sqrt 5 }} + \dfrac{1}{3}\arcsin \dfrac{1}{{\sqrt 5 }} + \dfrac{{2\pi k}}{3},\,\,\,\,\,\,\,\,\,}\\{x = \dfrac{\pi }{3}-\dfrac{1}{3}\arcsin \dfrac{2}{{\sqrt 5 }} + \dfrac{1}{3}\arcsin \dfrac{1}{{\sqrt 5 }} + \dfrac{{2\pi k}}{3},}\end{array}} \right.\,\,\,\,\,k \in Z.\)

Ответ:  \(\dfrac{1}{3}\arcsin \dfrac{2}{{\sqrt 5 }} + \dfrac{1}{3}\arcsin \dfrac{1}{{\sqrt 5 }} + \dfrac{{2\pi k}}{3};\,\,\,\,\,\dfrac{\pi }{3}-\dfrac{1}{3}\arcsin \dfrac{2}{{\sqrt 5 }} + \dfrac{1}{3}\arcsin \dfrac{1}{{\sqrt 5 }} + \dfrac{{2\pi k}}{3};\,\,\,\,\,k \in Z.\)