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If-x-is-a-positive-acute-angle-and-sinx-sin-2-x-sin-3-x-1-then-find-minimum-value-of-cot-2-x-




Question Number 216253 by MATHEMATICSAM last updated on 01/Feb/25
If x is a positive acute angle and  sinx + sin^2 x + sin^3 x = 1 then find  minimum value of cot^2 x.
Ifxisapositiveacuteangleandsinx+sin2x+sin3x=1thenfindminimumvalueofcot2x.
Commented by Frix last updated on 03/Feb/25
There′s exactly 1 solution to the given equation  thus there′s exactly one value of cot^2  x  sin x =((−1+(((17)/(27))−((√(33))/9))^(1/3) +(((17)/(27))+((√(33))/9))^(1/3) )/3)≈.543689  x≈32.9351°  cot^2  x =(2−((2(√(33)))/9))^(1/3) +(2+((2(√(33)))/9))^(1/3) ≈2.38298
Theresexactly1solutiontothegivenequationthustheresexactlyonevalueofcot2xsinx=1+(1727339)13+(1727+339)133.543689x32.9351°cot2x=(22339)13+(2+2339)132.38298
Answered by AntonCWX last updated on 03/Feb/25
let u=sin(x)  u+u^2 +u^3 =1  u^3 +u^2 +u−1=0    By Lagrange′s Resolvent,  ⇒z^2 +(2(1)^3 −9(1)(1)+27(−1))z+(1^2 −3(1))^3 =0  ⇒z^2 −34z−8=0  ⇒z=17∓3(√(33))    u=((−(1)+((17−3(√(33))))^(1/3) +((17+3(√(33))))^(1/3) )/3)=0.543689    sin(x)=0.543689⇒sin^2 (x)=0.295598  x=32.94°  cot^2 (x)=2.38209
letu=sin(x)u+u2+u3=1u3+u2+u1=0ByLagrangesResolvent,z2+(2(1)39(1)(1)+27(1))z+(123(1))3=0z234z8=0z=17333u=(1)+173333+17+33333=0.543689sin(x)=0.543689sin2(x)=0.295598x=32.94°cot2(x)=2.38209

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