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4cos-2-x-sin-x-2sin-2-x-3sin-x-where-pi-2-x-pi-2-

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Question Number 104904 by bemath last updated on 24/Jul/20
4cos^2 x sin x −2sin^2 x = 3sin x where −(π/2)≤x≤(π/2)
4cos2xsinx−2sin2x=3sinxwhere−π2⩽x⩽π2
Answered by bramlex last updated on 24/Jul/20
sin x (4cos^2 x−2sin x−3) = 0 { ((sin x = 0 → x_1 = 0)),((4cos^2 x −2sin x−3 = 0)) :} 4−4sin^2 x−2sin x−3 = 0 4sin^2 x+2sin x−1 = 0 sin x = ((−2 ± (√(4+16)))/8) = ((−1±(√5))/4) { ((x_2 = 18° = (π/(10)))),((x_3 = −54°= −((3π)/(10)) ⊸◊⊸)) :}
sinx(4cos2x−2sinx−3)=0{sinx=0→x1=04cos2x−2sinx−3=04−4sin2x−2sinx−3=04sin2x+2sinx−1=0sinx=−2±4+168=−1±54{x2=18°=π10x3=−54°=−3π10⊸◊⊸
Answered by Dwaipayan Shikari last updated on 24/Jul/20
2sinx(2cos^2 x−sinx)=3sinx 2−2sin^2 x−sinx=(3/2) or sinx=0 2sin^2 x+sinx−(1/2)=0 4sin^2 x+2sinx−1=0 sinx=((−2+(√(4+16)))/8)=((−(√5)±1)/4) or sinx=0 ⇒x=kπ sinx=sin(π/(10)) ( k∈Z) x=kπ±(π/(10)) or sinx=−((((√5)+1)/4)) x=−((3π)/(10)) Solutions are {(π/(10)),0,−((3π)/(10))}
2sinx(2cos2x−sinx)=3sinx2−2sin2x−sinx=32orsinx=02sin2x+sinx−12=04sin2x+2sinx−1=0sinx=−2+4+168=−5±14orsinx=0⇒x=kπsinx=sinπ10(k∈Z)x=kπ±π10orsinx=−(5+14)x=−3π10Solutionsare{π10,0,−3π10}
Answered by 1549442205PVT last updated on 24/Jul/20
⇔sinx(4cos^(2 ) x−2sinx−3)=0 ⇔sinx[(4(1−sin^(2 ) x)−2sinx−3]=0 ⇔sinx(4sin^2 x+2sinx−1)=0 i)sinx=0⇔x=kπ .Since x∈[−(π/2),(π/2)] ⇒x=0 ii)4sin^2 x+2sinx−1=0 ⇔sinx =((−1+(√5))/4) or sinx=((−1−(√5))/4) a)sinx=(((√5)−1)/4)=sin(π/(10))⇒x=(π/(10)) b)sinx=((−((√5)+1))/4)=sin((−3π)/(10))⇒x=((−3π)/(10)) Thus,x∈{0,(π/(10)),((−3π)/(10))}
⇔sinx(4cos2x−2sinx−3)=0⇔sinx[(4(1−sin2x)−2sinx−3]=0⇔sinx(4sin2x+2sinx−1)=0i)sinx=0⇔x=kπ.Sincex∈[−π2,π2]⇒x=0ii)4sin2x+2sinx−1=0⇔sinx=−1+54orsinx=−1−54a)sinx=5−14=sinπ10⇒x=π10b)sinx=−(5+1)4=sin−3π10⇒x=−3π10Thus,x∈{0,π10,−3π10}

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