Giải phương trình sau: (sin x + sin y + sin left( {x + y} right) = frac{{3sqrt 3 }}{2}).
Lời giải
Cách 1: Phương trình (sin x + sin y + sin left( {x + y} right) = frac{{3sqrt 3 }}{2},,,left( 1 right)).
Áp dụng bất đẳng thức Bunhiacốpxki và từ (left( 1 right)) ta có
(frac{{27}}{4} = {left( {frac{{3sqrt 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))
( = 3left[ {frac{{1 – cos 2x}}{2} + frac{{1 – cos 2y}}{2} + {{sin }^2}left( {x + y} right)} right])
( = 3left{ {1 – frac{1}{2}left[ {cos 2x + cos 2y} right] + left[ {1 – {{cos }^2}left( {x + y} right)} right]} right})
( = 3left[ {1 – cos left( {x + y} right).cos left( {x – y} right) + 1 – {{cos }^2}left( {x + y} right)} right])
( = 3left{ {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})
( = 3left{ {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 3left( {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} + k2pi \x = frac{{2pi }}{3} + k2pi end{array} right.\left[ begin{array}{l}y = frac{pi }{3} + n2pi \y = frac{pi }{3} + n2pi end{array} right.\sin left( {x + y} right) = frac{{sqrt 3 }}{2}end{array} right. Leftrightarrow left{ begin{array}{l}x = frac{pi }{3} + k2pi \y = frac{pi }{3} + n2pi 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)2pi } right] = 1,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,\cos left[ {frac{{2pi }}{3} + left( {k + n} right)2pi } right] + frac{1}{2}cos left[ {left( {k – n} right)2pi } right] = 0,,,end{array} right.)(luôn đúng với (forall ,k,,n in mathbb{Z})).
Vậy (left{ begin{array}{l}x = frac{pi }{3} + k2pi \y = frac{pi }{3} + n2pi end{array} right.left( {k,,n in mathbb{Z}} right)).
Cách 2: Phương trình (sin x + sin y + sin left( {x + y} right) = frac{{3sqrt 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{{3left( {{{sin }^2}x + {{cos }^2}x} right) + 3left( {{{sin }^2}y + {{cos }^2}y} right) + 3}}{{2sqrt 3 }} = frac{9}{{2sqrt 3 }} = frac{{3sqrt 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 xend{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} + k2pi \y = frac{pi }{3} + n2pi end{array} right.,,,,left( {k,,n in mathbb{Z}} right)).
Vậy (left{ begin{array}{l}x = frac{pi }{3} + k2pi \y = frac{pi }{3} + n2pi end{array} right.left( {k,,n in mathbb{Z}} right)).