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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
2cos2x−3cosx+sinx+1=0help
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=coss=sin2c2−3c+s+1=0;c2+s2=1⇒s2+c2=(−2c2+3c−1)2+c2=1⇔(4c4+9c2−12c3−6c+5c2)=0c(4c3−12c2+14c−6)=0⇒c(2c3−6c2+7c−3)=0c.(2c−2)(c2−2c+32)=0⇒c∈{0,1}c=0⇒sin(x)=−1⇒x=−π2+2kπc=1⇒sin(x)=0⇒x=2kπS={2kπ;−π2+2kπ};k∈Z
Commented by Ismoiljon_008 last updated on 17/Jul/24
thank you very much
thankyouverymuch
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}
2cosx2−3cosx+sinx+1=02t2−3t+1+1−t2=0=>(−2t2+3t−1)=1−t2=>4t4−12t3+4t2−6t+1+9t2=1−t2=>4t4−12t3+14t2−6t=0=>t(4t3−12t2+14t−6)=0=>t=0ou4t3−12t2+14t−6=0=>t(t−1)(4t2−8t+6)=0=>4t(t−1)(t2−2t+32)=0=>t=0out=1=>{cosx=0cosx=1=>{{x=π2+2kπx=−π2+2kπx=2kπSR={−π2+2kπ;π2+2kπ;+2kπ/k∈Z}

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