Giải phương trình sau: \(\sin x + \sin y + \sin \left( {x + y} \right) = \frac{{3\sqrt 3 }}{2}\).

Giải phương trình sau: \(\sin x + \sin y + \sin \left( {x + y} \right) = \frac{{3\sqrt 3 }}{2}\).

Lời giải

             

Cách 1: Phương trình \(\sin x + \sin y + \sin \left( {x + y} \right) = \frac{{3\sqrt 3 }}{2}\,\,\,\left( 1 \right)\).

Áp dụng bất đẳng thức Bunhiacốpxki và từ \(\left( 1 \right)\) ta có

\(\frac{{27}}{4} = {\left( {\frac{{3\sqrt 3 }}{2}} \right)^2} = {\left[ {\sin x + \sin y + \sin \left( {x + y} \right)} \right]^2} \le \left( {{1^2} + {1^2} + {1^2}} \right)\left( {{{\sin }^2}x + {{\sin }^2}y + {{\sin }^2}\left( {x + y} \right)} \right)\)

                     \( = 3\left[ {\frac{{1 – \cos 2x}}{2} + \frac{{1 – \cos 2y}}{2} + {{\sin }^2}\left( {x + y} \right)} \right]\)

 \( = 3\left\{ {1 – \frac{1}{2}\left[ {\cos 2x + \cos 2y} \right] + \left[ {1 – {{\cos }^2}\left( {x + y} \right)} \right]} \right\}\)

 \( = 3\left[ {1 – \cos \left( {x + y} \right).\cos \left( {x – y} \right) + 1 – {{\cos }^2}\left( {x + y} \right)} \right]\)

\( = 3\left\{ {2 – \left[ {{{\cos }^2}\left( {x + y} \right) + 2.\frac{1}{2}\cos \left( {x + y} \right).\cos \left( {x – y} \right) + \frac{1}{4}{{\cos }^2}\left( {x – y} \right)} \right] + \frac{1}{4}{{\cos }^2}\left( {x – y} \right)} \right\}\)

\( = 3\left\{ {2 – {{\left[ {\cos \left( {x + y} \right) + \frac{1}{2}\cos \left( {x – y} \right)} \right]}^2} + \frac{1}{4}{{\cos }^2}\left( {x – y} \right)} \right\}\)

\( \le 3\left( {2 – 0 + \frac{1}{4}} \right) = \frac{{27}}{4}\,\,\,\left( 2 \right)\) \(\left( {Do\,\,\,{{\cos }^2}\left( {x – y} \right) \le 1;\,{{\left[ {\cos \left( {x + y} \right) + \frac{1}{2}\cos \left( {x – y} \right)} \right]}^2} \ge 0} \right)\).

Từ \(\left( 2 \right)\) suy ra:

\(\left( 1 \right) \Leftrightarrow \left\{ \begin{array}{l}{\cos ^2}\left( {x – y} \right) = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 3 \right)\\\cos \left( {x + y} \right) + \frac{1}{2}\cos \left( {x – y} \right) = 0\,\,\,\,\,\,\,\,\,\,\,\,\left( 4 \right)\\\sin x = \sin y = \sin \left( {x + y} \right) = \frac{{\sqrt 3 }}{2}\,\,\,\,\,\,\,\,\,\left( 5 \right)\end{array} \right.\).

Giải \(\left( 5 \right)\): \(\sin x = \sin y = \sin \left( {x + y} \right) = \frac{{\sqrt 3 }}{2} \Leftrightarrow \left\{ \begin{array}{l}\sin x = \frac{{\sqrt 3 }}{2}\\\sin y = \frac{{\sqrt 3 }}{2}\\\sin \left( {x + y} \right) = \frac{{\sqrt 3 }}{2}\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}x = \frac{\pi }{3} + k2\pi \\x = \frac{{2\pi }}{3} + k2\pi \end{array} \right.\\\left[ \begin{array}{l}y = \frac{\pi }{3} + n2\pi \\y = \frac{\pi }{3} + n2\pi \end{array} \right.\\\sin \left( {x + y} \right) = \frac{{\sqrt 3 }}{2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = \frac{\pi }{3} + k2\pi \\y = \frac{\pi }{3} + n2\pi \end{array} \right.\left( {k,\,n \in \mathbb{Z}} \right)\,\,\,\,\left( * \right)\).

Thay \(\left( * \right)\) vào \(\left( 3 \right)\) và \(\left( 4 \right)\) ta được \(\left\{ \begin{array}{l}{\cos ^2}\left[ {\left( {k – n} \right)2\pi } \right] = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\\cos \left[ {\frac{{2\pi }}{3} + \left( {k + n} \right)2\pi } \right] + \frac{1}{2}\cos \left[ {\left( {k – n} \right)2\pi } \right] = 0\,\,\,\end{array} \right.\)(luôn đúng với \(\forall \,k,\,n \in \mathbb{Z}\)).

Vậy \(\left\{ \begin{array}{l}x = \frac{\pi }{3} + k2\pi \\y = \frac{\pi }{3} + n2\pi \end{array} \right.\left( {k,\,n \in \mathbb{Z}} \right)\).

Cách 2: Phương trình \(\sin x + \sin y + \sin \left( {x + y} \right) = \frac{{3\sqrt 3 }}{2}\,\,\,\left( 1 \right)\).

(Sử dụng bất đẳng thức: \(a.b \le \frac{{{a^2} + {b^2}}}{2},\,\forall a,\,b \in \mathbb{R}\), đẳng thức xảy ra \( \Leftrightarrow a = b\)).

Ta có

\(\sin x + \sin y + \sin \left( {x + y} \right)\)\( = \frac{2}{{\sqrt 3 }}\left( {\frac{{\sqrt 3 }}{2}.\sin x + \frac{{\sqrt 3 }}{2}.\sin y} \right) + \frac{1}{{\sqrt 3 }}\left( {\sqrt 3 .\cos y.\sin x + \sqrt 3 .\cos x.\sin y} \right)\)

\( \le \frac{2}{{\sqrt 3 }}.\frac{1}{2}\left( {\frac{3}{4} + {{\sin }^2}x + \frac{3}{4} + {{\sin }^2}y} \right) + \frac{1}{{\sqrt 3 }}.\frac{1}{2}\left( {3{{\cos }^2}y + {{\sin }^2}x + 3{{\cos }^2}x + {{\sin }^2}y} \right)\)

\( = \frac{{3\left( {{{\sin }^2}x + {{\cos }^2}x} \right) + 3\left( {{{\sin }^2}y + {{\cos }^2}y} \right) + 3}}{{2\sqrt 3 }} = \frac{9}{{2\sqrt 3 }} = \frac{{3\sqrt 3 }}{2}\).

Do đó, ta có \(\left( 1 \right) \Leftrightarrow \left\{ \begin{array}{l}\sin x = \sin y = \frac{{\sqrt 3 }}{2}\\\sin x = \sqrt 3 \cos y\\\sin y = \sqrt 3 \cos x\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\sin x = \sin y = \frac{{\sqrt 3 }}{2}\\\cos x = \cos y = \frac{1}{2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = \frac{\pi }{3} + k2\pi \\y = \frac{\pi }{3} + n2\pi \end{array} \right.\,\,\,\,\left( {k,\,n \in \mathbb{Z}} \right)\).

Vậy \(\left\{ \begin{array}{l}x = \frac{\pi }{3} + k2\pi \\y = \frac{\pi }{3} + n2\pi \end{array} \right.\left( {k,\,n \in \mathbb{Z}} \right)\).