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Question Number 102190 by dw last updated on 07/Jul/20

Find the value of x ∙∙[Trigonometric Subs.] x−(x/(√(1−x^2 )))=((91)/(60))

Findthevalueofx⋅⋅[TrigonometricSubs.]x−x1−x2=9160

Answered by bemath last updated on 07/Jul/20

((x((√(1−x^2 ))−1))/(√(1−x^2 ))) = ((91)/(60)) let x = sin θ ((sin θ.(cos θ−1))/(cos θ)) = ((91)/(60)) 60sin θcos θ−60sin θ=91cos θ

x(1−x2−1)1−x2=9160letx=sinθsinθ.(cosθ−1)cosθ=916060sinθcosθ−60sinθ=91cosθ

Answered by 1549442205 last updated on 07/Jul/20

Put x=cosϕ(0≤ϕ≤π)we have cosϕ−((cosϕ)/(sinϕ))=((91)/(60))⇔60cosϕsinϕ−60cosϕ=91sinϕ(1) Put t=tan(ϕ/2).We get (1)⇔60((2(1−t^2 )t)/((1+t^2 )^2 ))−60((1−t^2 )/(1+t^2 ))=91((2t)/(1+t^2 )) ⇔120(t−t^3 )−60(1−t^4 )=182(t^3 +t) ⇔30t^4 −151t^3 −31t−30=0 ⇔30t^4 −151t^3 −31t−30=0 we get two roots(by Calculator) t_1 =−0.45781901488,t_2 =5.08098312713 From this we get ϕ_1 =2arctan t_1 =−49°11′54 (is rejected as ϕ_1 outside [0;π]) ϕ_2 =2arctan t_2 =157°43′53 Hence,x_2 =cosϕ_2 =−0.9254187445 Thus,the unique root x=−0.925417445 satisfying our prpblem

Putx=cosφ(0⩽φ⩽π)wehavecosφ−cosφsinφ=9160⇔60cosφsinφ−60cosφ=91sinφ(1)Putt=tanφ2.Weget(1)⇔602(1−t2)t(1+t2)2−601−t21+t2=912t1+t2⇔120(t−t3)−60(1−t4)=182(t3+t)⇔30t4−151t3−31t−30=0⇔30t4−151t3−31t−30=0wegettworoots(byCalculator)t1=−0.45781901488,t2=5.08098312713Fromthiswegetφ1=2arctant1=−49°11′54(isrejectedasφ1outside[0;π])φ2=2arctant2=157°43′53Hence,x2=cosφ2=−0.9254187445Thus,theuniquerootx=−0.925417445satisfyingourprpblem

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