Задача 8

Решить уравнение

\(\sqrt {{\rm{co}}{{\rm{s}}^2}x + \frac{1}{2}} + \sqrt {{\rm{si}}{{\rm{n}}^2}x + \frac{1}{2}} = 2.\)


Решение

Пусть

\(\left\{ {\begin{array}{*{20}{c}}{\sqrt {{\rm{co}}{{\rm{s}}^2}x + \frac{1}{2}} = u > 0,}\\{\sqrt {{\rm{si}}{{\rm{n}}^2}x + \frac{1}{2}} = v > 0;}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{{\rm{co}}{{\rm{s}}^2}x + \frac{1}{2} = {u^2} > 0,}\\{{\rm{si}}{{\rm{n}}^2}x + \frac{1}{2} = {v^2} > 0;}\end{array}} \right. \Rightarrow {u^2} + {v^2} = 2.\)

Получили следующую систему

\(\left\{ {\begin{array}{*{20}{c}}{u + v = 2,}\\{{u^2} + {v^2} = 2;}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{u + v = 2,}\\{{{\left( {u + v} \right)}^2} - 2uv = 2;}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{u + v = 2,}\\{uv = 1;}\end{array}} \right. \Rightarrow u = 1,\;v = 1.\)

Тогда

\(\sqrt {{\rm{co}}{{\rm{s}}^2}x + \frac{1}{2}} = 1,\;\;{\rm{co}}{{\rm{s}}^2}x + \frac{1}{2} = 1,\;\;{\rm{co}}{{\rm{s}}^2}x = \frac{1}{2},\;\;\cos x = \pm \frac{{\sqrt 2 }}{2},\)

\(x = \frac{\pi }{4} + \frac{{\pi k}}{2},\;\;k \in \mathbb{Z}.\)

Ответ: \(x = \frac{\pi }{4} + \frac{{\pi k}}{2},\;\;k \in \mathbb{Z}\)