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2cos-2-x-3cosx-sinx-1-0-help-

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Question Number 209656 by Ismoiljon_008 last updated on 17/Jul/24
2cos^2 x−3cosx+sinx+1=0 help
$$ \\ $$$$\:\:\:\mathrm{2}{cos}^{\mathrm{2}} {x}−\mathrm{3}{cosx}+{sinx}+\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:{help} \\ $$$$ \\ $$
Answered by Berbere last updated on 17/Jul/24
c=cos s=sin 2c^2 −3c+s+1=0;c^2 +s^2 =1⇒ s^2 +c^2 =(−2c^2 +3c−1)^2 +c^2 =1 ⇔(4c^4 +9c^2 −12c^3 −6c+5c^2 )=0 c(4c^3 −12c^2 +14c−6)=0 ⇒c(2c^3 −6c^2 +7c−3)=0 c.(2c−2)(c^2 −2c+(3/2))=0⇒c∈{0,1} c=0⇒ sin(x)=−1⇒x=−(π/2)+2kπ c=1⇒sin(x)=0⇒x=2kπ S={2kπ;−(π/2)+2kπ};k∈Z
$${c}={cos} \\ $$$${s}={sin} \\ $$$$\mathrm{2}{c}^{\mathrm{2}} −\mathrm{3}{c}+{s}+\mathrm{1}=\mathrm{0};{c}^{\mathrm{2}} +{s}^{\mathrm{2}} =\mathrm{1}\Rightarrow \\ $$$${s}^{\mathrm{2}} +{c}^{\mathrm{2}} =\left(−\mathrm{2}{c}^{\mathrm{2}} +\mathrm{3}{c}−\mathrm{1}\right)^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{1} \\ $$$$\Leftrightarrow\left(\mathrm{4}{c}^{\mathrm{4}} +\mathrm{9}{c}^{\mathrm{2}} −\mathrm{12}{c}^{\mathrm{3}} −\mathrm{6}{c}+\mathrm{5}{c}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${c}\left(\mathrm{4}{c}^{\mathrm{3}} −\mathrm{12}{c}^{\mathrm{2}} +\mathrm{14}{c}−\mathrm{6}\right)=\mathrm{0} \\ $$$$\Rightarrow{c}\left(\mathrm{2}{c}^{\mathrm{3}} −\mathrm{6}{c}^{\mathrm{2}} +\mathrm{7}{c}−\mathrm{3}\right)=\mathrm{0} \\ $$$${c}.\left(\mathrm{2}{c}−\mathrm{2}\right)\left({c}^{\mathrm{2}} −\mathrm{2}{c}+\frac{\mathrm{3}}{\mathrm{2}}\right)=\mathrm{0}\Rightarrow{c}\in\left\{\mathrm{0},\mathrm{1}\right\} \\ $$$${c}=\mathrm{0}\Rightarrow\:{sin}\left({x}\right)=−\mathrm{1}\Rightarrow{x}=−\frac{\pi}{\mathrm{2}}+\mathrm{2}{k}\pi\: \\ $$$${c}=\mathrm{1}\Rightarrow{sin}\left({x}\right)=\mathrm{0}\Rightarrow{x}=\mathrm{2}{k}\pi \\ $$$${S}=\left\{\mathrm{2}{k}\pi;−\frac{\pi}{\mathrm{2}}+\mathrm{2}{k}\pi\right\};{k}\in\mathbb{Z} \\ $$
Commented by Ismoiljon_008 last updated on 17/Jul/24
thank you very much
$$\:\:\:{thank}\:{you}\:{very}\:{much}\: \\ $$$$ \\ $$
Answered by lepuissantcedricjunior last updated on 18/Jul/24
2cos^2 x−3cosx+sinx+1=0 2t^2 −3t+1+(√(1−t^2 ))=0 =>(−2t^2 +3t−1)=(√(1−t^2 )) =>4t^4 −12t^3 +4t^2 −6t+1+9t^2 =1−t^2 =>4t^4 −12t^3 +14t^2 −6t=0 =>t(4t^3 −12t^2 +14t−6)=0 =>t=0ou 4t^3 −12t^2 +14t−6=0 =>t(t−1)(4t^2 −8t+6)=0 =>4t(t−1)(t^2 −2t+(3/2))=0 =>t=0ou t=1 => { ((cosx=0)),((cosx=1)) :}=> { ( { ((x=(𝛑/2)+2k𝛑)),((x=−(𝛑/2)+2k𝛑)) :}),((x=2k𝛑)) :} S_R ={−(𝛑/2)+2k𝛑;(𝛑/2)+2k𝛑;+2k𝛑/k∈Z}
$$\mathrm{2}\boldsymbol{{co}}\overset{\mathrm{2}} {\boldsymbol{{s}x}}−\mathrm{3}\boldsymbol{{cosx}}+\boldsymbol{{sinx}}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{2}\boldsymbol{{t}}^{\mathrm{2}} −\mathrm{3}\boldsymbol{{t}}+\mathrm{1}+\sqrt{\mathrm{1}−\boldsymbol{{t}}^{\mathrm{2}} }=\mathrm{0} \\ $$$$=>\left(−\mathrm{2}\boldsymbol{{t}}^{\mathrm{2}} +\mathrm{3}\boldsymbol{{t}}−\mathrm{1}\right)=\sqrt{\mathrm{1}−\boldsymbol{{t}}^{\mathrm{2}} } \\ $$$$=>\mathrm{4}\boldsymbol{{t}}^{\mathrm{4}} −\mathrm{12}\boldsymbol{{t}}^{\mathrm{3}} +\mathrm{4}\boldsymbol{{t}}^{\mathrm{2}} −\mathrm{6}\boldsymbol{{t}}+\mathrm{1}+\mathrm{9}\boldsymbol{{t}}^{\mathrm{2}} =\mathrm{1}−\boldsymbol{{t}}^{\mathrm{2}} \\ $$$$=>\mathrm{4}\boldsymbol{{t}}^{\mathrm{4}} −\mathrm{12}\boldsymbol{{t}}^{\mathrm{3}} +\mathrm{14}\boldsymbol{{t}}^{\mathrm{2}} −\mathrm{6}\boldsymbol{{t}}=\mathrm{0} \\ $$$$=>\boldsymbol{{t}}\left(\mathrm{4}\boldsymbol{{t}}^{\mathrm{3}} −\mathrm{12}\boldsymbol{{t}}^{\mathrm{2}} +\mathrm{14}\boldsymbol{{t}}−\mathrm{6}\right)=\mathrm{0} \\ $$$$=>\boldsymbol{{t}}=\mathrm{0}\boldsymbol{{ou}}\:\mathrm{4}\boldsymbol{{t}}^{\mathrm{3}} −\mathrm{12}\boldsymbol{{t}}^{\mathrm{2}} +\mathrm{14}\boldsymbol{{t}}−\mathrm{6}=\mathrm{0} \\ $$$$=>\boldsymbol{{t}}\left(\boldsymbol{{t}}−\mathrm{1}\right)\left(\mathrm{4}\boldsymbol{{t}}^{\mathrm{2}} −\mathrm{8}\boldsymbol{{t}}+\mathrm{6}\right)=\mathrm{0} \\ $$$$=>\mathrm{4}\boldsymbol{{t}}\left(\boldsymbol{{t}}−\mathrm{1}\right)\left(\boldsymbol{{t}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{t}}+\frac{\mathrm{3}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$$=>\boldsymbol{{t}}=\mathrm{0}\boldsymbol{{ou}}\:\boldsymbol{{t}}=\mathrm{1} \\ $$$$=>\begin{cases}{\boldsymbol{{cosx}}=\mathrm{0}}\\{\boldsymbol{{cosx}}=\mathrm{1}}\end{cases}=>\begin{cases}{\begin{cases}{\boldsymbol{{x}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}+\mathrm{2}\boldsymbol{{k}\pi}}\\{\boldsymbol{{x}}=−\frac{\boldsymbol{\pi}}{\mathrm{2}}+\mathrm{2}\boldsymbol{{k}\pi}}\end{cases}}\\{\boldsymbol{{x}}=\mathrm{2}\boldsymbol{{k}\pi}}\end{cases} \\ $$$$\boldsymbol{{S}}_{\mathbb{R}} =\left\{−\frac{\boldsymbol{\pi}}{\mathrm{2}}+\mathrm{2}\boldsymbol{{k}\pi};\frac{\boldsymbol{\pi}}{\mathrm{2}}+\mathrm{2}\boldsymbol{{k}\pi};+\mathrm{2}\boldsymbol{{k}\pi}/\boldsymbol{{k}}\in\mathbb{Z}\right\} \\ $$

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