Уравнения и неравенства, содержащие обратные тригонометрические функции. Задача 5

ОТВЕТ: \(-\dfrac{1}{3}.\)

\(\arcsin \dfrac{{\sqrt {3x + 2} }}{2}-\dfrac{\pi }{2} + {\rm{arctg}}\sqrt {\dfrac{2}{{x + 1}}}  = 0.\)

Запишем ОДЗ:  \(\left\{ {\begin{array}{*{20}{c}}{3x + 2 \ge 0,\,\,\,}\\{\dfrac{{\sqrt {3x + 2} }}{2} \le 1,}\\{\dfrac{2}{{x + 1}} \ge 0\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x \ge -\dfrac{2}{3},}\\{x \le \dfrac{2}{3},}\\{x > -1}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x \in \left[ {-\dfrac{2}{3};\,\dfrac{2}{3}} \right].\)

\(\arcsin \dfrac{{\sqrt {3x + 2} }}{2}-\dfrac{\pi }{2} + {\rm{arctg}}\sqrt {\dfrac{2}{{x + 1}}}  = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\dfrac{\pi }{2}-\arcsin \dfrac{{\sqrt {3x + 2} }}{2} = {\rm{arctg}}\sqrt {\dfrac{2}{{x + 1}}} ;\)

\({\rm{tg}}\left( {\dfrac{\pi }{2}-\arcsin \dfrac{{\sqrt {3x + 2} }}{2}} \right) = {\rm{tg}}\left( {{\rm{arctg}}\sqrt {\dfrac{2}{{x + 1}}} } \right)\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{\rm{ctg}}\left( {\arcsin \dfrac{{\sqrt {3x + 2} }}{2}} \right) = \sqrt {\dfrac{2}{{x + 1}}} \,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\dfrac{{\cos \left( {\arcsin \dfrac{{\sqrt {3x + 2} }}{2}} \right)}}{{\sin \left( {\arcsin \dfrac{{\sqrt {3x + 2} }}{2}} \right)}} = \sqrt {\dfrac{2}{{x + 1}}} \,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\dfrac{{\sqrt {1-\dfrac{{3x + 2}}{4}} }}{{\dfrac{{\sqrt {3x + 2} }}{2}}} = \sqrt {\dfrac{2}{{x + 1}}} \,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\sqrt {\dfrac{{2-3x}}{{3x + 2}}}  = \sqrt {\dfrac{2}{{x + 1}}} ;\)

\(\dfrac{{2-3x}}{{3x + 2}} = \dfrac{2}{{x + 1}};\,\,\,\,\,\,3{x^2} + 7x + 2 = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = -\dfrac{1}{3},}\\{x = -2.\,\,}\end{array}} \right.\)

Корень  \(x = -2\)  не удовлетворяет условию  \(x\, \in \,\left[ {-\dfrac{2}{3};\dfrac{2}{3}} \right]\).

Проверкой убеждаемся, что  \(x = -\dfrac{1}{3}\)  является корнем исходного уравнения.

Ответ:  \(-\dfrac{1}{3}\).