Уравнение вида a sin x + b cos x = c. Задача 14

ОТВЕТ: \(\arcsin \dfrac{7}{{29}} + \arcsin \dfrac{{21}}{{29}} + 2\pi k;\quad \pi -\arcsin \dfrac{7}{{29}} + \arcsin \dfrac{{21}}{{29}} + 2\pi k;\quad \,k \in Z.\)

\(20\sin x-21\cos x = 7.\)

Уравнение вида:  \(a\sin x + b\cos x = c,\;\;\)где \(\;a = 20,\,\,\,\,b = -21,\;\;c = 7.\)

Разделим обе части исходного уравнения на  \(\sqrt {{a^2} + {b^2}} ,\;\) то есть на  \(\sqrt {{{20}^2} + {{\left( {-21} \right)}^2}}  = 29.\)

\(\dfrac{{20}}{{29}}\sin x-\dfrac{{21}}{{29}}\cos x = \dfrac{7}{{29}}.\)

Так как  \({\left( {\dfrac{{20}}{{29}}} \right)^2} + {\left( {\dfrac{{21}}{{29}}} \right)^2} = 1,\)  то пусть  \(\cos \varphi  = \dfrac{{20}}{{29}},\,\,\,\sin \varphi  = \dfrac{{21}}{{29}}.\)  Тогда уравнение примет вид:

\(\dfrac{{20}}{{29}}\sin x-\dfrac{{21}}{{29}}\cos x = \dfrac{7}{{29}}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\cos \varphi \sin x-\sin \varphi \cos x = \dfrac{7}{{29}}\,\,\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\,\,\sin \left( {x-\varphi } \right) = \dfrac{7}{{29}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x-\varphi  = \arcsin \dfrac{7}{{29}} + 2\pi k,\,\,\,\,\,\,}\\{x-\varphi  = \pi -\arcsin \dfrac{7}{{29}} + 2\pi k}\end{array}} \right.\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \arcsin \dfrac{7}{{29}} + \varphi  + 2\pi k,\,\,\,\,\,\,\,\,\,}\\{x = \pi -\arcsin \dfrac{7}{{29}} + \varphi  + 2\pi k,}\end{array}} \right.\,\,\,\,k \in Z.\)

Так как  \(\sin \varphi  = \dfrac{{21}}{{29}},\)  то   \(\varphi  = \arcsin \dfrac{{21}}{{29}}\)  и

\(\left[ {\begin{array}{*{20}{c}}{x = \arcsin \dfrac{7}{{29}} + \arcsin \dfrac{{21}}{{29}} + 2\pi k,\,\,\,\,\,\,\,\,\,}\\{x = \pi -\arcsin \dfrac{7}{{29}} + \arcsin \dfrac{{21}}{{29}} + 2\pi k,}\end{array}} \right.\,\,\,\,\,k \in Z.\)

Ответ:  \(\arcsin \dfrac{7}{{29}} + \arcsin \dfrac{{21}}{{29}} + 2\pi k;\quad \pi -\arcsin \dfrac{7}{{29}} + \arcsin \dfrac{{21}}{{29}} + 2\pi k;\quad \,k \in Z.\)