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Solve-using-trig-method-8x-3-4x-2-4x-1-0-




Question Number 131421 by Lordose last updated on 04/Feb/21
Solve using trig method    8x^3 −4x^2 −4x+1=0
Solveusingtrigmethod8x34x24x+1=0
Commented by Dwaipayan Shikari last updated on 04/Feb/21
x=y+(1/6)  8y^3 +(1/(27))+4y^2 +((2y)/3)−4y^2 −((4y)/3)−(1/9)−4y−(2/3)+1=0  ⇒8y^3 −((14y)/3)+(7/(27))=0  y=Rcosθ    ⇒   cos^3 θ−(7/(12R^2 ))cosθ+(7/(216R^3 ))=0       ((27)/( 64(√7)))  Comparing with cos^3 θ−(3/4)cosθ−(1/4)cos3θ=0  (7/(12R^2 ))=(3/4)⇒R=±((√7)/3)         cos3θ=±((27)/(64(√7)))  θ=kπ+(1/3)cos^(−1) ±(((27)/(64(√7)))) ⇒Rcosθ=((±(√7))/3)cos(kπ+(1/3)cos^(−1) (±((27)/(64(√7)))))  x=±((√7)/3)cos(kπ+(1/3)cos^(−1) (±((27)/(64(√7)))))−(1/6)
x=y+168y3+127+4y2+2y34y24y3194y23+1=08y314y3+727=0y=Rcosθcos3θ712R2cosθ+7216R3=027647Comparingwithcos3θ34cosθ14cos3θ=0712R2=34R=±73cos3θ=±27647θ=kπ+13cos1±(27647)Rcosθ=±73cos(kπ+13cos1(±27647))x=±73cos(kπ+13cos1(±27647))16