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

Russian olympiad find real solution of the system { ((sin x+2sin (x+y+z)=0)),((sin y+3sin (x+y+z)=0)),((sin z+4sin (x+y+z)=0)) :}

$${Russian}\:{olympiad}\: \\ $$$${find}\:{real}\:{solution}\:{of}\:{the}\:{system}\: \\ $$$$\begin{cases}{\mathrm{sin}\:{x}+\mathrm{2sin}\:\left({x}+{y}+{z}\right)=\mathrm{0}}\\{\mathrm{sin}\:{y}+\mathrm{3sin}\:\left({x}+{y}+{z}\right)=\mathrm{0}}\\{\mathrm{sin}\:{z}+\mathrm{4sin}\:\left({x}+{y}+{z}\right)=\mathrm{0}}\end{cases} \\ $$$$ \\ $$

Answered by benjo_mathlover last updated on 25/Oct/20

→ sin (x+y+z) = −((sin x)/2)=−((sin y)/3)=−((sin z)/4) let sin x = k then { ((sin y=(3/2)k)),((sin z=2k)) :} →sin ((x+y)+z)=sin (x+y)cos z+cos (x+y)sin z = cos z(sin xcos y+cos xsin y)+sin z(cos xcos y−sin xsin y) =sin xcos ycos z+sin ycos xcos z+sin zcos xcos y−sin xsin ysin z tobe continue..

$$\rightarrow\:\mathrm{sin}\:\left({x}+{y}+{z}\right)\:=\:−\frac{\mathrm{sin}\:{x}}{\mathrm{2}}=−\frac{\mathrm{sin}\:{y}}{\mathrm{3}}=−\frac{\mathrm{sin}\:{z}}{\mathrm{4}} \\ $$$${let}\:\mathrm{sin}\:{x}\:=\:{k}\:{then}\:\begin{cases}{\mathrm{sin}\:{y}=\frac{\mathrm{3}}{\mathrm{2}}{k}}\\{\mathrm{sin}\:{z}=\mathrm{2}{k}}\end{cases} \\ $$$$\rightarrow\mathrm{sin}\:\left(\left({x}+{y}\right)+{z}\right)=\mathrm{sin}\:\left({x}+{y}\right)\mathrm{cos}\:{z}+\mathrm{cos}\:\left({x}+{y}\right)\mathrm{sin}\:{z} \\ $$$$=\:\mathrm{cos}\:{z}\left(\mathrm{sin}\:{x}\mathrm{cos}\:{y}+\mathrm{cos}\:{x}\mathrm{sin}\:{y}\right)+\mathrm{sin}\:{z}\left(\mathrm{cos}\:{x}\mathrm{cos}\:{y}−\mathrm{sin}\:{x}\mathrm{sin}\:{y}\right) \\ $$$$=\mathrm{sin}\:{x}\mathrm{cos}\:{y}\mathrm{cos}\:{z}+\mathrm{sin}\:{y}\mathrm{cos}\:{x}\mathrm{cos}\:{z}+\mathrm{sin}\:{z}\mathrm{cos}\:{x}\mathrm{cos}\:{y}−\mathrm{sin}\:{x}\mathrm{sin}\:{y}\mathrm{sin}\:{z} \\ $$$${tobe}\:{continue}.. \\ $$

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

{ ((sin x+2sin (x+y+z)=0(1))),((sin y+3sin (x+y+z)=0(2))),((sin z+4sin (x+y+z)=0(3))) :}(∗) ⇒((sinx)/(−2))=((siny)/(−3))=((sinz)/(−4))=−m ⇔∣4m∣=∣sinz∣≤1⇒((−1)/4)≤m≤(1/4)(∗∗) ⇒sinx=2m,siny=3m,sinz=4m cosx=±(√(1−4m^2 )),cosy=±(√(1−9m^2 )) cosz=±(√(1−16m^2 )) (4) •Consider case cosx,cosy,cosz>0 sin(x+y+z)=sinxcos(y+z)+cosxsin(y+z) =sinx(cosycosz−sinysinz) +cosx(sinycosz+cosysinz) =2m(√(1−25m^2 +144m^4 ))−24m^3 +3m(√(1−20m^2 +64m^4 )) +4m(√(1−13m^2 +36m^4 )) Replace into (3)we get: 4m+8m(√(1−25m^2 +144m^4 ))−48m^3 +12m(√(1−20m^2 +64m^4 )) +16m(√(1−13m^2 +36m^4 ))=0 i)m=0⇒sinx=siny=sinz=0 ⇒x=mπ,y=nπ,z=kπ ii)4+8(√(1−25m^2 +144m^4 ))−96m^2 +12(√(1−20m^2 +64m^4 )) +16(√(1−13m^2 +36m^4 ))=0 =4−96m^2 +8(√((1−25m^2 −144m^4 ))) +12(√((1−20m^2 +64)) +16(√((((13)/(12))−6m^2 )−((25)/(144)))) ≥4−96.(1/(16))+16(√((((13)/(12))−(6/(16)))^2 −((25)/(144)))) =−2+16(√((21)/(64)))=−2+2(√(21))>0 Since m^2 ≤(1/(16)), hence L.H.S>0 ∀m satisfying (∗∗),so the equation has no roots Thus,the system of equations (∗)has a family of solutions are: (x,y,z)=(l𝛑,n𝛑,k𝛑) (with l,n,k∈Z) •The cosx>0,cosy,cosz<0.We have (3)−ii)⇔4+8(√(1−25m^2 +144m^4 )) −96m^2 −12(√(1−20m^2 +64m^4 )) −16(√(1−13m^2 +36m^4 )) ⇔8(√(1−25m^2 +144m^4 )) =96m^2 −4+ 12(√(1−20m^2 +64m^4 )) +16(√(1−13m^2 +36m^4 ))(♠) We have 8(√(1−25m^2 +144m^4 )) ≤ 12(√(1−20m^2 +64m^4 )) ⇔2(√(1−25m^2 +144m^4 )) ≤3(√(1−20m^2 +64m^4 ))⇔ 4(1−25m^2 +144m^4 )≤9(1−20m^2 +64m^4 ) ⇔80m^2 ≤5⇔m^2 ≤16 (1)This inequality is true due to (∗∗).On the other hands, by above proof we have: 16(√(1−13m^2 +36m^2 ))>2(√(21))>4⇒ 96m^2 −4+16(√(1−13m^2 +36m^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)⇔4−8(√(1−25m^2 +144m^4 ))−96m^2 −12(√(1−20m^2 +64m^4 )) +16(√(1−13m^2 +36m^4 )) ⇔2(√(1−25m^2 +144m^4 )) +3(√(1−20m^2 +64m^4 )) =4(√(1−13m^2 +36m^4 ))−24m^2 +1⇔ 4(1−25m^2 +144m^4 )+9(1−20m^2 +64m^4 ) +12(√(9216m^8 −4480m^6 +708m^4 −45m^2 +1)) =16(1−13m^2 +36m^4 )+576m^4 −48m^2 +1 +8(1−24m^2 )(√(1−13m^2 +36m^4 )) ⇔12(√(9216m^8 −4480m^6 +708m^4 −45m^2 +1)) =24m^2 +4+8(1−25m^2 )(√(1−13m^2 +36m^4 )) ⇒144(9216m^8 −4480m^6 +708m^4 −45m^2 +1) =576m^4 +16+192m^2 +64(1−50m^2 +625m^4 )(1−13m^2 +36m^4 ) +16(4+24m^2 )(1−25m^2 )(√(1−13m^2 +3m^4 )) ⇔−112896m^8 −9920m^6 +59115m^4 −2640m^2 +64=16(4+24m^2 )(1−25m^2 )(√(1−13m^2 +36m^4 )) be continued

$$\begin{cases}{\mathrm{sin}\:{x}+\mathrm{2sin}\:\left({x}+{y}+{z}\right)=\mathrm{0}\left(\mathrm{1}\right)}\\{\mathrm{sin}\:{y}+\mathrm{3sin}\:\left({x}+{y}+{z}\right)=\mathrm{0}\left(\mathrm{2}\right)}\\{\mathrm{sin}\:{z}+\mathrm{4sin}\:\left({x}+{y}+{z}\right)=\mathrm{0}\left(\mathrm{3}\right)}\end{cases}\left(\ast\right)\: \\ $$$$\Rightarrow\frac{\mathrm{sinx}}{−\mathrm{2}}=\frac{\mathrm{siny}}{−\mathrm{3}}=\frac{\mathrm{sinz}}{−\mathrm{4}}=−\mathrm{m} \\ $$$$\Leftrightarrow\mid\mathrm{4m}\mid=\mid\mathrm{sinz}\mid\leqslant\mathrm{1}\Rightarrow\frac{−\mathrm{1}}{\mathrm{4}}\leqslant\mathrm{m}\leqslant\frac{\mathrm{1}}{\mathrm{4}}\left(\ast\ast\right) \\ $$$$\Rightarrow\mathrm{sinx}=\mathrm{2m},\mathrm{siny}=\mathrm{3m},\mathrm{sinz}=\mathrm{4m} \\ $$$$\mathrm{cosx}=\pm\sqrt{\mathrm{1}−\mathrm{4m}^{\mathrm{2}} },\mathrm{cosy}=\pm\sqrt{\mathrm{1}−\mathrm{9m}^{\mathrm{2}} } \\ $$$$\mathrm{cosz}=\pm\sqrt{\mathrm{1}−\mathrm{16m}^{\mathrm{2}} \:}\:\left(\mathrm{4}\right) \\ $$$$\bullet\mathrm{Consider}\:\mathrm{case}\:\mathrm{cosx},\mathrm{cosy},\mathrm{cosz}>\mathrm{0} \\ $$$$\mathrm{sin}\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)=\mathrm{sinxcos}\left(\mathrm{y}+\mathrm{z}\right)+\mathrm{cosxsin}\left(\mathrm{y}+\mathrm{z}\right) \\ $$$$=\mathrm{sinx}\left(\mathrm{cosycosz}−\mathrm{sinysinz}\right) \\ $$$$+\mathrm{cosx}\left(\mathrm{sinycosz}+\mathrm{cosysinz}\right) \\ $$$$=\mathrm{2m}\sqrt{\mathrm{1}−\mathrm{25m}^{\mathrm{2}} +\mathrm{144m}^{\mathrm{4}} }−\mathrm{24m}^{\mathrm{3}} \\ $$$$+\mathrm{3m}\sqrt{\mathrm{1}−\mathrm{20m}^{\mathrm{2}} +\mathrm{64m}^{\mathrm{4}} }\:+\mathrm{4m}\sqrt{\mathrm{1}−\mathrm{13m}^{\mathrm{2}} +\mathrm{36m}^{\mathrm{4}} } \\ $$$$\mathrm{Replace}\:\mathrm{into}\:\left(\mathrm{3}\right)\mathrm{we}\:\mathrm{get}: \\ $$$$\mathrm{4m}+\mathrm{8m}\sqrt{\mathrm{1}−\mathrm{25m}^{\mathrm{2}} +\mathrm{144m}^{\mathrm{4}} }−\mathrm{48m}^{\mathrm{3}} \\ $$$$+\mathrm{12m}\sqrt{\mathrm{1}−\mathrm{20m}^{\mathrm{2}} +\mathrm{64m}^{\mathrm{4}} }\:+\mathrm{16m}\sqrt{\mathrm{1}−\mathrm{13m}^{\mathrm{2}} +\mathrm{36m}^{\mathrm{4}} }=\mathrm{0} \\ $$$$\left.\mathrm{i}\right)\mathrm{m}=\mathrm{0}\Rightarrow\mathrm{sinx}=\mathrm{siny}=\mathrm{sinz}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{m}\pi,\mathrm{y}=\mathrm{n}\pi,\mathrm{z}=\mathrm{k}\pi \\ $$$$\left.\mathrm{ii}\right)\mathrm{4}+\mathrm{8}\sqrt{\mathrm{1}−\mathrm{25m}^{\mathrm{2}} +\mathrm{144m}^{\mathrm{4}} }−\mathrm{96m}^{\mathrm{2}} \\ $$$$+\mathrm{12}\sqrt{\mathrm{1}−\mathrm{20m}^{\mathrm{2}} +\mathrm{64m}^{\mathrm{4}} }\:+\mathrm{16}\sqrt{\mathrm{1}−\mathrm{13m}^{\mathrm{2}} +\mathrm{36m}^{\mathrm{4}} }=\mathrm{0} \\ $$$$=\mathrm{4}−\mathrm{96m}^{\mathrm{2}} +\mathrm{8}\sqrt{\left(\mathrm{1}−\mathrm{25m}^{\mathrm{2}} −\mathrm{144m}^{\mathrm{4}} \right)} \\ $$$$+\mathrm{12}\sqrt{\left(\mathrm{1}−\mathrm{20m}^{\mathrm{2}} +\mathrm{64}\right.}\:+\mathrm{16}\sqrt{\left(\frac{\mathrm{13}}{\mathrm{12}}−\mathrm{6m}^{\mathrm{2}} \right)−\frac{\mathrm{25}}{\mathrm{144}}} \\ $$$$\geqslant\mathrm{4}−\mathrm{96}.\frac{\mathrm{1}}{\mathrm{16}}+\mathrm{16}\sqrt{\left(\frac{\mathrm{13}}{\mathrm{12}}−\frac{\mathrm{6}}{\mathrm{16}}\right)^{\mathrm{2}} −\frac{\mathrm{25}}{\mathrm{144}}} \\ $$$$=−\mathrm{2}+\mathrm{16}\sqrt{\frac{\mathrm{21}}{\mathrm{64}}}=−\mathrm{2}+\mathrm{2}\sqrt{\mathrm{21}}>\mathrm{0} \\ $$$$\mathrm{Since}\:\mathrm{m}^{\mathrm{2}} \leqslant\frac{\mathrm{1}}{\mathrm{16}}, \\ $$$$\mathrm{hence}\:\mathrm{L}.\mathrm{H}.\mathrm{S}>\mathrm{0}\:\forall\mathrm{m}\:\mathrm{satisfying}\:\left(\ast\ast\right),\mathrm{so} \\ $$$$\mathrm{the}\:\mathrm{equation}\:\mathrm{has}\:\mathrm{no}\:\mathrm{roots} \\ $$$$\mathrm{Thus},\mathrm{the}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equations}\:\left(\ast\right)\mathrm{has} \\ $$$$\mathrm{a}\:\mathrm{family}\:\mathrm{of}\:\mathrm{solutions}\:\mathrm{are}: \\ $$$$\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}},\boldsymbol{\mathrm{z}}\right)=\left(\mathrm{l}\boldsymbol{\pi},\boldsymbol{\mathrm{n}\pi},\boldsymbol{\mathrm{k}\pi}\right)\:\left(\boldsymbol{\mathrm{with}}\:\mathrm{l},\boldsymbol{\mathrm{n}},\boldsymbol{\mathrm{k}}\in\boldsymbol{\mathrm{Z}}\right) \\ $$$$\bullet\mathrm{The}\:\mathrm{cosx}>\mathrm{0},\mathrm{cosy},\mathrm{cosz}<\mathrm{0}.\mathrm{We}\:\mathrm{have} \\ $$$$\left.\left(\mathrm{3}\right)−\mathrm{ii}\right)\Leftrightarrow\mathrm{4}+\mathrm{8}\sqrt{\mathrm{1}−\mathrm{25m}^{\mathrm{2}} +\mathrm{144m}^{\mathrm{4}} }\:−\mathrm{96m}^{\mathrm{2}} \\ $$$$−\mathrm{12}\sqrt{\mathrm{1}−\mathrm{20m}^{\mathrm{2}} +\mathrm{64m}^{\mathrm{4}} }\:−\mathrm{16}\sqrt{\mathrm{1}−\mathrm{13m}^{\mathrm{2}} +\mathrm{36m}^{\mathrm{4}} } \\ $$$$\Leftrightarrow\mathrm{8}\sqrt{\mathrm{1}−\mathrm{25m}^{\mathrm{2}} +\mathrm{144m}^{\mathrm{4}} }\:=\mathrm{96m}^{\mathrm{2}} −\mathrm{4}+ \\ $$$$\mathrm{12}\sqrt{\mathrm{1}−\mathrm{20m}^{\mathrm{2}} +\mathrm{64m}^{\mathrm{4}} }\:+\mathrm{16}\sqrt{\mathrm{1}−\mathrm{13m}^{\mathrm{2}} +\mathrm{36m}^{\mathrm{4}} }\left(\spadesuit\right) \\ $$$$\mathrm{We}\:\mathrm{have}\:\mathrm{8}\sqrt{\mathrm{1}−\mathrm{25m}^{\mathrm{2}} +\mathrm{144m}^{\mathrm{4}} }\:\leqslant \\ $$$$\mathrm{12}\sqrt{\mathrm{1}−\mathrm{20m}^{\mathrm{2}} +\mathrm{64m}^{\mathrm{4}} }\:\Leftrightarrow\mathrm{2}\sqrt{\mathrm{1}−\mathrm{25m}^{\mathrm{2}} +\mathrm{144m}^{\mathrm{4}} }\: \\ $$$$\leqslant\mathrm{3}\sqrt{\mathrm{1}−\mathrm{20m}^{\mathrm{2}} +\mathrm{64m}^{\mathrm{4}} }\Leftrightarrow \\ $$$$\mathrm{4}\left(\mathrm{1}−\mathrm{25m}^{\mathrm{2}} +\mathrm{144m}^{\mathrm{4}} \right)\leqslant\mathrm{9}\left(\mathrm{1}−\mathrm{20m}^{\mathrm{2}} +\mathrm{64m}^{\mathrm{4}} \right) \\ $$$$\Leftrightarrow\mathrm{80m}^{\mathrm{2}} \leqslant\mathrm{5}\Leftrightarrow\mathrm{m}^{\mathrm{2}} \leqslant\mathrm{16}\:\left(\mathrm{1}\right)\mathrm{This}\:\mathrm{inequality}\: \\ $$$$\mathrm{is}\:\mathrm{true}\:\mathrm{due}\:\mathrm{to}\:\left(\ast\ast\right).\mathrm{On}\:\mathrm{the}\:\mathrm{other}\:\mathrm{hands}, \\ $$$$\mathrm{by}\:\mathrm{above}\:\mathrm{proof}\:\mathrm{we}\:\mathrm{have}: \\ $$$$\mathrm{16}\sqrt{\mathrm{1}−\mathrm{13m}^{\mathrm{2}} +\mathrm{36m}^{\mathrm{2}} }>\mathrm{2}\sqrt{\mathrm{21}}>\mathrm{4}\Rightarrow \\ $$$$\mathrm{96m}^{\mathrm{2}} −\mathrm{4}+\mathrm{16}\sqrt{\mathrm{1}−\mathrm{13m}^{\mathrm{2}} +\mathrm{36m}^{\mathrm{2}} }>\mathrm{0}\left(\mathrm{2}\right) \\ $$$$\mathrm{From}\:\left(\mathrm{1}\right)\left(\mathrm{2}\right)\:\mathrm{we}\:\mathrm{infer}\:\mathrm{that}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\left(\spadesuit\right)\:\mathrm{has}\:\mathrm{no}\:\mathrm{root}.\mathrm{Hence}\:\mathrm{this}\:\mathrm{case}\:\mathrm{proved} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\:\mathrm{equation}\:\mathrm{has}\:\mathrm{no}\:\mathrm{roots} \\ $$$$\bullet\mathrm{The}\:\mathrm{cosx},\mathrm{cosy}<\mathrm{0},\mathrm{cosz}>\mathrm{0}.\mathrm{We}\:\mathrm{have} \\ $$$$\left.\left(\mathrm{3}\right)−\mathrm{ii}\right)\Leftrightarrow\mathrm{4}−\mathrm{8}\sqrt{\mathrm{1}−\mathrm{25m}^{\mathrm{2}} +\mathrm{144m}^{\mathrm{4}} }−\mathrm{96m}^{\mathrm{2}} \\ $$$$−\mathrm{12}\sqrt{\mathrm{1}−\mathrm{20m}^{\mathrm{2}} +\mathrm{64m}^{\mathrm{4}} }\:+\mathrm{16}\sqrt{\mathrm{1}−\mathrm{13m}^{\mathrm{2}} +\mathrm{36m}^{\mathrm{4}} } \\ $$$$\Leftrightarrow\mathrm{2}\sqrt{\mathrm{1}−\mathrm{25m}^{\mathrm{2}} +\mathrm{144m}^{\mathrm{4}} }\:+\mathrm{3}\sqrt{\mathrm{1}−\mathrm{20m}^{\mathrm{2}} +\mathrm{64m}^{\mathrm{4}} } \\ $$$$=\mathrm{4}\sqrt{\mathrm{1}−\mathrm{13m}^{\mathrm{2}} +\mathrm{36m}^{\mathrm{4}} }−\mathrm{24m}^{\mathrm{2}} +\mathrm{1}\Leftrightarrow \\ $$$$\mathrm{4}\left(\mathrm{1}−\mathrm{25m}^{\mathrm{2}} +\mathrm{144m}^{\mathrm{4}} \right)+\mathrm{9}\left(\mathrm{1}−\mathrm{20m}^{\mathrm{2}} +\mathrm{64m}^{\mathrm{4}} \right) \\ $$$$+\mathrm{12}\sqrt{\mathrm{9216m}^{\mathrm{8}} −\mathrm{4480m}^{\mathrm{6}} +\mathrm{708m}^{\mathrm{4}} −\mathrm{45m}^{\mathrm{2}} +\mathrm{1}} \\ $$$$=\mathrm{16}\left(\mathrm{1}−\mathrm{13m}^{\mathrm{2}} +\mathrm{36m}^{\mathrm{4}} \right)+\mathrm{576m}^{\mathrm{4}} −\mathrm{48m}^{\mathrm{2}} +\mathrm{1} \\ $$$$+\mathrm{8}\left(\mathrm{1}−\mathrm{24m}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−\mathrm{13m}^{\mathrm{2}} +\mathrm{36m}^{\mathrm{4}} } \\ $$$$\Leftrightarrow\mathrm{12}\sqrt{\mathrm{9216m}^{\mathrm{8}} −\mathrm{4480m}^{\mathrm{6}} +\mathrm{708m}^{\mathrm{4}} −\mathrm{45m}^{\mathrm{2}} +\mathrm{1}} \\ $$$$=\mathrm{24m}^{\mathrm{2}} +\mathrm{4}+\mathrm{8}\left(\mathrm{1}−\mathrm{25m}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−\mathrm{13m}^{\mathrm{2}} +\mathrm{36m}^{\mathrm{4}} } \\ $$$$\Rightarrow\mathrm{144}\left(\mathrm{9216m}^{\mathrm{8}} −\mathrm{4480m}^{\mathrm{6}} +\mathrm{708m}^{\mathrm{4}} −\mathrm{45m}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$=\mathrm{576m}^{\mathrm{4}} +\mathrm{16}+\mathrm{192m}^{\mathrm{2}} +\mathrm{64}\left(\mathrm{1}−\mathrm{50m}^{\mathrm{2}} +\mathrm{625m}^{\mathrm{4}} \right)\left(\mathrm{1}−\mathrm{13m}^{\mathrm{2}} +\mathrm{36m}^{\mathrm{4}} \right) \\ $$$$+\mathrm{16}\left(\mathrm{4}+\mathrm{24m}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{25m}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−\mathrm{13m}^{\mathrm{2}} +\mathrm{3m}^{\mathrm{4}} } \\ $$$$\Leftrightarrow−\mathrm{112896m}^{\mathrm{8}} −\mathrm{9920m}^{\mathrm{6}} +\mathrm{59115m}^{\mathrm{4}} \\ $$$$−\mathrm{2640m}^{\mathrm{2}} +\mathrm{64}=\mathrm{16}\left(\mathrm{4}+\mathrm{24m}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{25m}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−\mathrm{13m}^{\mathrm{2}} +\mathrm{36m}^{\mathrm{4}} } \\ $$$$\mathrm{be}\:\mathrm{continued} \\ $$

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

sin(x+y+z)=cosxcosycosz(tanx+tany+tanz−tanxtanytanz) sin(x+y+z)=((sinx)/(−2))=((siny)/(−3))=((sinz)/(−4))=a so a=(1−4a^2 )^(1/2) (1−9a^2 )^(1/2) (1−16a^2 )^(1/2) (((−2a)/( (√(1−4a^2 ))))+((−3a)/( (√(1−9a^2 ))))+((−4a)/( (√(1−16a^2 )) ))+((24a^3 )/( (√((1−4a^2 )(1−9a^2 )(1−16a^2 )))))) to solve...

$${sin}\left({x}+{y}+{z}\right)={cosxcosycosz}\left({tanx}+{tany}+{tanz}−{tanxtanytanz}\right) \\ $$$${sin}\left({x}+{y}+{z}\right)=\frac{{sinx}}{−\mathrm{2}}=\frac{{siny}}{−\mathrm{3}}=\frac{{sinz}}{−\mathrm{4}}={a} \\ $$$${so} \\ $$$${a}=\left(\mathrm{1}−\mathrm{4}{a}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−\mathrm{9}{a}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−\mathrm{16}{a}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\frac{−\mathrm{2}{a}}{\:\sqrt{\mathrm{1}−\mathrm{4}{a}^{\mathrm{2}} }}+\frac{−\mathrm{3}{a}}{\:\sqrt{\mathrm{1}−\mathrm{9}{a}^{\mathrm{2}} }}+\frac{−\mathrm{4}{a}}{\:\sqrt{\mathrm{1}−\mathrm{16}{a}^{\mathrm{2}} }\:}+\frac{\mathrm{24}{a}^{\mathrm{3}} }{\:\sqrt{\left(\mathrm{1}−\mathrm{4}{a}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{9}{a}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{16}{a}^{\mathrm{2}} \right)}}\right) \\ $$$${to}\:{solve}... \\ $$

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