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Question Number 119494 by liberty last updated on 25/Oct/20

R u s s i a n o l y m p i a d

f i n d r e a l s o l u t i o n o f t h e s y s t e m

{\begin{cases} \sin x + 2 \sin (x + y + z) = 0 \\ \sin y + 3 \sin (x + y + z) = 0 \\ \sin z + 4 \sin (x + y + z) = 0 \end{cases}

Answered by benjo_mathlover last updated on 25/Oct/20

\to \sin (x + y + z) = - \frac{\sin x}{2} = - \frac{\sin y}{3} = - \frac{\sin z}{4}

l e t \sin x = k t h e n {\begin{cases} \sin y = \frac{3}{2} k \\ \sin z = 2 k \end{cases}

\to \sin ((x + y) + z) = \sin (x + y) \cos z + \cos (x + y) \sin z

= \cos z (\sin x \cos y + \cos x \sin y) + \sin z (\cos x \cos y - \sin x \sin y)

= \sin x \cos y \cos z + \sin y \cos x \cos z + \sin z \cos x \cos y - \sin x \sin y \sin z

t o b e c o n t i n u e . .

Answered by 1549442205PVT last updated on 25/Oct/20

{\begin{cases} \sin x + 2 \sin (x + y + z) = 0 (1) \\ \sin y + 3 \sin (x + y + z) = 0 (2) \\ \sin z + 4 \sin (x + y + z) = 0 (3) \end{cases} (*)

\Rightarrow \frac{sinx}{- 2} = \frac{siny}{- 3} = \frac{sinz}{- 4} = - m

\Leftrightarrow∣ 4 m ∣=∣ sinz ∣⩽ 1 \Rightarrow \frac{- 1}{4} ⩽ m ⩽ \frac{1}{4} (* *)

\Rightarrow sinx = 2 m, siny = 3 m, sinz = 4 m

cosx = \pm \sqrt{1 - {4 m}^{2}}, cosy = \pm \sqrt{1 - {9 m}^{2}}

cosz = \pm \sqrt{1 - {16 m}^{2}} (4)

∙ Consider case cosx, cosy, cosz > 0

\sin (x + y + z) = sinxcos (y + z) + cosxsin (y + z)

= sinx (cosycosz - sinysinz)

+ cosx (sinycosz + cosysinz)

= 2 m \sqrt{1 - {25 m}^{2} + {144 m}^{4}} - {24 m}^{3}

+ 3 m \sqrt{1 - {20 m}^{2} + {64 m}^{4}} + 4 m \sqrt{1 - {13 m}^{2} + {36 m}^{4}}

Replace into (3) we get :

4 m + 8 m \sqrt{1 - {25 m}^{2} + {144 m}^{4}} - {48 m}^{3}

+ 12 m \sqrt{1 - {20 m}^{2} + {64 m}^{4}} + 16 m \sqrt{1 - {13 m}^{2} + {36 m}^{4}} = 0

i) m = 0 \Rightarrow sinx = siny = sinz = 0

\Rightarrow x = m π, y = n π, z = k π

ii) 4 + 8 \sqrt{1 - {25 m}^{2} + {144 m}^{4}} - {96 m}^{2}

+ 12 \sqrt{1 - {20 m}^{2} + {64 m}^{4}} + 16 \sqrt{1 - {13 m}^{2} + {36 m}^{4}} = 0

= 4 - {96 m}^{2} + 8 \sqrt{(1 - {25 m}^{2} - {144 m}^{4})}

+ 12 \sqrt{(1 - {20 m}^{2} + 64} + 16 \sqrt{(\frac{13}{12} - {6 m}^{2}) - \frac{25}{144}}

⩾ 4 - 96 . \frac{1}{16} + 16 \sqrt{{(\frac{13}{12} - \frac{6}{16})}^{2} - \frac{25}{144}}

= - 2 + 16 \sqrt{\frac{21}{64}} = - 2 + 2 \sqrt{21} > 0

Since m^{2} ⩽ \frac{1}{16},

hence L . H . S > 0 \forall m satisfying (* *), so

the equation has no roots

Thus, the system of equations (*) has

a family of solutions are :

(x, y, z) = (l π, n π, k π) (with l, n, k \in Z)

∙ The cosx > 0, cosy, cosz < 0 . We have

(3) - ii) \Leftrightarrow 4 + 8 \sqrt{1 - {25 m}^{2} + {144 m}^{4}} - {96 m}^{2}

- 12 \sqrt{1 - {20 m}^{2} + {64 m}^{4}} - 16 \sqrt{1 - {13 m}^{2} + {36 m}^{4}}

\Leftrightarrow 8 \sqrt{1 - {25 m}^{2} + {144 m}^{4}} = {96 m}^{2} - 4 +

12 \sqrt{1 - {20 m}^{2} + {64 m}^{4}} + 16 \sqrt{1 - {13 m}^{2} + {36 m}^{4}} (♠)

We have 8 \sqrt{1 - {25 m}^{2} + {144 m}^{4}} ⩽

12 \sqrt{1 - {20 m}^{2} + {64 m}^{4}} \Leftrightarrow 2 \sqrt{1 - {25 m}^{2} + {144 m}^{4}}

⩽ 3 \sqrt{1 - {20 m}^{2} + {64 m}^{4}} \Leftrightarrow

4 (1 - {25 m}^{2} + {144 m}^{4}) ⩽ 9 (1 - {20 m}^{2} + {64 m}^{4})

\Leftrightarrow {80 m}^{2} ⩽ 5 \Leftrightarrow m^{2} ⩽ 16 (1) This inequality

is true due to (* *) . On the other hands,

by above proof we have :

16 \sqrt{1 - {13 m}^{2} + {36 m}^{2}} > 2 \sqrt{21} > 4 \Rightarrow

{96 m}^{2} - 4 + 16 \sqrt{1 - {13 m}^{2} + {36 m}^{2}} > 0 (2)

From (1) (2) we infer that the equation

(♠) has no root . Hence this case proved

equation has no roots

∙ The cosx, cosy < 0, cosz > 0 . We have

(3) - ii) \Leftrightarrow 4 - 8 \sqrt{1 - {25 m}^{2} + {144 m}^{4}} - {96 m}^{2}

- 12 \sqrt{1 - {20 m}^{2} + {64 m}^{4}} + 16 \sqrt{1 - {13 m}^{2} + {36 m}^{4}}

\Leftrightarrow 2 \sqrt{1 - {25 m}^{2} + {144 m}^{4}} + 3 \sqrt{1 - {20 m}^{2} + {64 m}^{4}}

= 4 \sqrt{1 - {13 m}^{2} + {36 m}^{4}} - {24 m}^{2} + 1 \Leftrightarrow

4 (1 - {25 m}^{2} + {144 m}^{4}) + 9 (1 - {20 m}^{2} + {64 m}^{4})

+ 12 \sqrt{{9216 m}^{8} - {4480 m}^{6} + {708 m}^{4} - {45 m}^{2} + 1}

= 16 (1 - {13 m}^{2} + {36 m}^{4}) + {576 m}^{4} - {48 m}^{2} + 1

+ 8 (1 - {24 m}^{2}) \sqrt{1 - {13 m}^{2} + {36 m}^{4}}

\Leftrightarrow 12 \sqrt{{9216 m}^{8} - {4480 m}^{6} + {708 m}^{4} - {45 m}^{2} + 1}

= {24 m}^{2} + 4 + 8 (1 - {25 m}^{2}) \sqrt{1 - {13 m}^{2} + {36 m}^{4}}

\Rightarrow 144 ({9216 m}^{8} - {4480 m}^{6} + {708 m}^{4} - {45 m}^{2} + 1)

= {576 m}^{4} + 16 + {192 m}^{2} + 64 (1 - {50 m}^{2} + {625 m}^{4}) (1 - {13 m}^{2} + {36 m}^{4})

+ 16 (4 + {24 m}^{2}) (1 - {25 m}^{2}) \sqrt{1 - {13 m}^{2} + {3 m}^{4}}

\Leftrightarrow - {112896 m}^{8} - {9920 m}^{6} + {59115 m}^{4}

- {2640 m}^{2} + 64 = 16 (4 + {24 m}^{2}) (1 - {25 m}^{2}) \sqrt{1 - {13 m}^{2} + {36 m}^{4}}

be continued

Answered by TANMAY PANACEA last updated on 25/Oct/20

s i n (x + y + z) = c o s x c o s y c o s z (t a n x + t a n y + t a n z - t a n x t a n y t a n z)

s i n (x + y + z) = \frac{s i n x}{- 2} = \frac{s i n y}{- 3} = \frac{s i n z}{- 4} = a

s o

a = {(1 - 4 a^{2})}^{\frac{1}{2}} {(1 - 9 a^{2})}^{\frac{1}{2}} {(1 - 16 a^{2})}^{\frac{1}{2}} (\frac{- 2 a}{\sqrt{1 - 4 a^{2}}} + \frac{- 3 a}{\sqrt{1 - 9 a^{2}}} + \frac{- 4 a}{\sqrt{1 - 16 a^{2}}} + \frac{24 a^{3}}{\sqrt{(1 - 4 a^{2}) (1 - 9 a^{2}) (1 - 16 a^{2})}})

t o s o l v e \dots