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Question-200873




Question Number 200873 by cortano12 last updated on 25/Nov/23
Commented by witcher3 last updated on 26/Nov/23
tan(((3x)/2))∈[−(√3),(√3)]  tg(3.(x/2))=((3tg((x/2))−tg^3 ((x/2)))/(1−3tg^2 ((x/2))))  sin(x)(√(3−tan^2 (((3x)/2))))=2+cos(x)  ⇔(√(3−[((3tan((x/2))−tan^3 ((x/2)))/(1−3tan^2 ((x/2))))]^2 ))=((3cos^2 ((x/2))+sin^2 ((x/2)))/(2sin((x/2))cos((x/2))))=((3+tg^2 ((x/2)))/(2tg((x/2))))  a=tan((x/2))  ⇔(√(3−[(((3a−a^3 ))/((1−3a^2 )^2 ))]^2 ))=((3+a^2 )/(2a))  ⇒((3(9a^4 −6a^2 +1)−(9a^2 +a^6 −6a^4 ))/((1−3a^2 )^2 ))=(((a^4 +6a^2 +9))/(4a^2 ))  a^2 =y  ⇔((−y^3 +33y^2 −27y+3)/((1−3y)^2 ))=((y^2 +6y+9)/(4y))  4y(−y^3 +33y^2 −27y+3)=(9y^2 −6y+1)(y^2 +6y+9)  9y^4 +48y^3 +46y^2 −48y+9=−4y^4 +132y^3 −108y^2 +12y  13y^4 −84y^3 +154y^2 −60y+9=0...P  In Z/3Z ⇒2y^2 ≡0[3]⇒3∣y  y=3 is solution of P  13y^4 −84y^3 +154y^2 −60y+9  =(y−3)^2 (13y^2 −6y+1)  only one solution overR  a=+_− (√3)  tan((x/2))=(√3)⇒(x/2)=tan^(−1) ((√3))⇒x=((2π)/3)+2kπ..worck  tan((x/2))=−(√3)⇒x=−((2π)/3)+2kπ...not worcking
$$\mathrm{tan}\left(\frac{\mathrm{3x}}{\mathrm{2}}\right)\in\left[−\sqrt{\mathrm{3}},\sqrt{\mathrm{3}}\right] \\ $$$$\mathrm{tg}\left(\mathrm{3}.\frac{\mathrm{x}}{\mathrm{2}}\right)=\frac{\mathrm{3tg}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)−\mathrm{tg}^{\mathrm{3}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)}{\mathrm{1}−\mathrm{3tg}^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)} \\ $$$$\mathrm{sin}\left(\mathrm{x}\right)\sqrt{\mathrm{3}−\mathrm{tan}^{\mathrm{2}} \left(\frac{\mathrm{3x}}{\mathrm{2}}\right)}=\mathrm{2}+\mathrm{cos}\left(\mathrm{x}\right) \\ $$$$\Leftrightarrow\sqrt{\mathrm{3}−\left[\frac{\mathrm{3tan}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)−\mathrm{tan}^{\mathrm{3}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)}{\mathrm{1}−\mathrm{3tan}^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)}\right]^{\mathrm{2}} }=\frac{\mathrm{3cos}^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)+\mathrm{sin}^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)}{\mathrm{2sin}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\mathrm{cos}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)}=\frac{\mathrm{3}+\mathrm{tg}^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)}{\mathrm{2tg}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)} \\ $$$$\mathrm{a}=\mathrm{tan}\left(\frac{\mathrm{x}}{\mathrm{2}}\right) \\ $$$$\Leftrightarrow\sqrt{\mathrm{3}−\left[\frac{\left(\mathrm{3a}−\mathrm{a}^{\mathrm{3}} \right)}{\left(\mathrm{1}−\mathrm{3a}^{\mathrm{2}} \right)^{\mathrm{2}} }\right]^{\mathrm{2}} }=\frac{\mathrm{3}+\mathrm{a}^{\mathrm{2}} }{\mathrm{2a}} \\ $$$$\Rightarrow\frac{\mathrm{3}\left(\mathrm{9a}^{\mathrm{4}} −\mathrm{6a}^{\mathrm{2}} +\mathrm{1}\right)−\left(\mathrm{9a}^{\mathrm{2}} +\mathrm{a}^{\mathrm{6}} −\mathrm{6a}^{\mathrm{4}} \right)}{\left(\mathrm{1}−\mathrm{3a}^{\mathrm{2}} \right)^{\mathrm{2}} }=\frac{\left(\mathrm{a}^{\mathrm{4}} +\mathrm{6a}^{\mathrm{2}} +\mathrm{9}\right)}{\mathrm{4a}^{\mathrm{2}} } \\ $$$$\mathrm{a}^{\mathrm{2}} =\mathrm{y} \\ $$$$\Leftrightarrow\frac{−\mathrm{y}^{\mathrm{3}} +\mathrm{33y}^{\mathrm{2}} −\mathrm{27y}+\mathrm{3}}{\left(\mathrm{1}−\mathrm{3y}\right)^{\mathrm{2}} }=\frac{\mathrm{y}^{\mathrm{2}} +\mathrm{6y}+\mathrm{9}}{\mathrm{4y}} \\ $$$$\mathrm{4y}\left(−\mathrm{y}^{\mathrm{3}} +\mathrm{33y}^{\mathrm{2}} −\mathrm{27y}+\mathrm{3}\right)=\left(\mathrm{9y}^{\mathrm{2}} −\mathrm{6y}+\mathrm{1}\right)\left(\mathrm{y}^{\mathrm{2}} +\mathrm{6y}+\mathrm{9}\right) \\ $$$$\mathrm{9y}^{\mathrm{4}} +\mathrm{48y}^{\mathrm{3}} +\mathrm{46y}^{\mathrm{2}} −\mathrm{48y}+\mathrm{9}=−\mathrm{4y}^{\mathrm{4}} +\mathrm{132y}^{\mathrm{3}} −\mathrm{108y}^{\mathrm{2}} +\mathrm{12y} \\ $$$$\mathrm{13y}^{\mathrm{4}} −\mathrm{84y}^{\mathrm{3}} +\mathrm{154y}^{\mathrm{2}} −\mathrm{60y}+\mathrm{9}=\mathrm{0}…\mathrm{P} \\ $$$$\mathrm{In}\:\mathbb{Z}/\mathrm{3}\mathbb{Z}\:\Rightarrow\mathrm{2y}^{\mathrm{2}} \equiv\mathrm{0}\left[\mathrm{3}\right]\Rightarrow\mathrm{3}\mid\mathrm{y} \\ $$$$\mathrm{y}=\mathrm{3}\:\mathrm{is}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{P} \\ $$$$\mathrm{13y}^{\mathrm{4}} −\mathrm{84y}^{\mathrm{3}} +\mathrm{154y}^{\mathrm{2}} −\mathrm{60y}+\mathrm{9} \\ $$$$=\left(\mathrm{y}−\mathrm{3}\right)^{\mathrm{2}} \left(\mathrm{13y}^{\mathrm{2}} −\mathrm{6y}+\mathrm{1}\right) \\ $$$$\mathrm{only}\:\mathrm{one}\:\mathrm{solution}\:\mathrm{over}\mathbb{R} \\ $$$$\mathrm{a}=\underset{−} {+}\sqrt{\mathrm{3}} \\ $$$$\mathrm{tan}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\sqrt{\mathrm{3}}\Rightarrow\frac{\mathrm{x}}{\mathrm{2}}=\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{3}}\right)\Rightarrow\mathrm{x}=\frac{\mathrm{2}\pi}{\mathrm{3}}+\mathrm{2k}\pi..\mathrm{worck} \\ $$$$\mathrm{tan}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)=−\sqrt{\mathrm{3}}\Rightarrow\mathrm{x}=−\frac{\mathrm{2}\pi}{\mathrm{3}}+\mathrm{2k}\pi…\mathrm{not}\:\mathrm{worcking} \\ $$$$ \\ $$$$ \\ $$
Answered by Frix last updated on 26/Nov/23
c=cos x  (√((1−c)(1+c)))(√(3−(((1−c)(1+2c)^2 )/((1+c)(1−2c)^2 ))))−c=2  c≠−1∧c≠(1/2)  (√(1−c))(√(2(8c^3 −6c+1)))=(2+c)∣1−2c∣  c^4 −(c^3 /5)−((11c^2 )/(20))+(c/(10))+(1/(10))=0  (c+(1/2))^2 (c^2 −((6c)/5)+(2/5))=0  c=−(1/2)  cos x =−(1/2)  x=2nπ±((2π)/3)  Testing ⇒ x=2nπ+((2π)/3)
$${c}=\mathrm{cos}\:{x} \\ $$$$\sqrt{\left(\mathrm{1}−{c}\right)\left(\mathrm{1}+{c}\right)}\sqrt{\mathrm{3}−\frac{\left(\mathrm{1}−{c}\right)\left(\mathrm{1}+\mathrm{2}{c}\right)^{\mathrm{2}} }{\left(\mathrm{1}+{c}\right)\left(\mathrm{1}−\mathrm{2}{c}\right)^{\mathrm{2}} }}−{c}=\mathrm{2} \\ $$$${c}\neq−\mathrm{1}\wedge{c}\neq\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\sqrt{\mathrm{1}−{c}}\sqrt{\mathrm{2}\left(\mathrm{8}{c}^{\mathrm{3}} −\mathrm{6}{c}+\mathrm{1}\right)}=\left(\mathrm{2}+{c}\right)\mid\mathrm{1}−\mathrm{2}{c}\mid \\ $$$${c}^{\mathrm{4}} −\frac{{c}^{\mathrm{3}} }{\mathrm{5}}−\frac{\mathrm{11}{c}^{\mathrm{2}} }{\mathrm{20}}+\frac{{c}}{\mathrm{10}}+\frac{\mathrm{1}}{\mathrm{10}}=\mathrm{0} \\ $$$$\left({c}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \left({c}^{\mathrm{2}} −\frac{\mathrm{6}{c}}{\mathrm{5}}+\frac{\mathrm{2}}{\mathrm{5}}\right)=\mathrm{0} \\ $$$${c}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{cos}\:{x}\:=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}=\mathrm{2}{n}\pi\pm\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\mathrm{Testing}\:\Rightarrow\:{x}=\mathrm{2}{n}\pi+\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$

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