Question-157790

Question Number 157790 by MathSh last updated on 28/Oct/21

Answered by MJS_new last updated on 28/Oct/21

let c=cos x ((32c^5 −40c^3 +9c))^(1/3) +((16c^5 −20c^3 +4c))^(1/3) =(c)^(1/3) u+v=w ∣^3 u^3 +3u^2 v+3uv^2 +v^3 =w^3 u^3 +3uv(u+v)+v^3 =w^3 u^3 +3uvw+v^3 =w^3 3uvw=w^3 −(u^3 +v^3 ) 27u^3 v^3 w^3 =(w^3 −(u^3 +v^3 ))^3 inserting we get 108c^3 (4c^4 −5c^2 +1)(32c^4 −40c^2 +9)=−1728c^3 (4c^4 −5c^2 +1)^3 ⇒ c=0∨4c^4 −5c^2 +1=0 32c^4 −40c^2 +9=−16(4c^4 −5c^2 +1)^2 let c^2 =d+(5/8) 32d^2 −(7/2)=−(((64d^2 −9)^2 )/(16)) d^4 −(5/(32))d^2 +((25)/(4096))=0 (d^2 −(5/(64)))^2 =0 d=±((√5)/8) c^2 =((5±(√5))/8) now we have c=0 c=±1 c=±(1/2) c=±(√((5−(√5))/8)) c=±(√((5+(√5))/8)) testing leads to c=0∨c=±(1/2)∨c=±1 for 0≤x<2π we get x∈{0, 2, 3, 4, 6, 8, 9, 10}×(π/6)

$$\mathrm{let}\:{c}=\mathrm{cos}\:{x} \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{32}{c}^{\mathrm{5}} −\mathrm{40}{c}^{\mathrm{3}} +\mathrm{9}{c}}+\sqrt[{\mathrm{3}}]{\mathrm{16}{c}^{\mathrm{5}} −\mathrm{20}{c}^{\mathrm{3}} +\mathrm{4}{c}}=\sqrt[{\mathrm{3}}]{{c}} \\ $$$${u}+{v}={w}\:\:\:\:\:\mid^{\mathrm{3}} \\ $$$${u}^{\mathrm{3}} +\mathrm{3}{u}^{\mathrm{2}} {v}+\mathrm{3}{uv}^{\mathrm{2}} +{v}^{\mathrm{3}} ={w}^{\mathrm{3}} \\ $$$${u}^{\mathrm{3}} +\mathrm{3}{uv}\left({u}+{v}\right)+{v}^{\mathrm{3}} ={w}^{\mathrm{3}} \\ $$$${u}^{\mathrm{3}} +\mathrm{3}{uvw}+{v}^{\mathrm{3}} ={w}^{\mathrm{3}} \\ $$$$\mathrm{3}{uvw}={w}^{\mathrm{3}} −\left({u}^{\mathrm{3}} +{v}^{\mathrm{3}} \right) \\ $$$$\mathrm{27}{u}^{\mathrm{3}} {v}^{\mathrm{3}} {w}^{\mathrm{3}} =\left({w}^{\mathrm{3}} −\left({u}^{\mathrm{3}} +{v}^{\mathrm{3}} \right)\right)^{\mathrm{3}} \\ $$$$\mathrm{inserting}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{108}{c}^{\mathrm{3}} \left(\mathrm{4}{c}^{\mathrm{4}} −\mathrm{5}{c}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{32}{c}^{\mathrm{4}} −\mathrm{40}{c}^{\mathrm{2}} +\mathrm{9}\right)=−\mathrm{1728}{c}^{\mathrm{3}} \left(\mathrm{4}{c}^{\mathrm{4}} −\mathrm{5}{c}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} \\ $$$$\Rightarrow\:{c}=\mathrm{0}\vee\mathrm{4}{c}^{\mathrm{4}} −\mathrm{5}{c}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{32}{c}^{\mathrm{4}} −\mathrm{40}{c}^{\mathrm{2}} +\mathrm{9}=−\mathrm{16}\left(\mathrm{4}{c}^{\mathrm{4}} −\mathrm{5}{c}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{let}\:{c}^{\mathrm{2}} ={d}+\frac{\mathrm{5}}{\mathrm{8}} \\ $$$$\mathrm{32}{d}^{\mathrm{2}} −\frac{\mathrm{7}}{\mathrm{2}}=−\frac{\left(\mathrm{64}{d}^{\mathrm{2}} −\mathrm{9}\right)^{\mathrm{2}} }{\mathrm{16}} \\ $$$${d}^{\mathrm{4}} −\frac{\mathrm{5}}{\mathrm{32}}{d}^{\mathrm{2}} +\frac{\mathrm{25}}{\mathrm{4096}}=\mathrm{0} \\ $$$$\left({d}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{64}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${d}=\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{8}} \\ $$$${c}^{\mathrm{2}} =\frac{\mathrm{5}\pm\sqrt{\mathrm{5}}}{\mathrm{8}} \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{have} \\ $$$${c}=\mathrm{0} \\ $$$${c}=\pm\mathrm{1} \\ $$$${c}=\pm\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${c}=\pm\sqrt{\frac{\mathrm{5}−\sqrt{\mathrm{5}}}{\mathrm{8}}} \\ $$$${c}=\pm\sqrt{\frac{\mathrm{5}+\sqrt{\mathrm{5}}}{\mathrm{8}}} \\ $$$$\mathrm{testing}\:\mathrm{leads}\:\mathrm{to} \\ $$$${c}=\mathrm{0}\vee{c}=\pm\frac{\mathrm{1}}{\mathrm{2}}\vee{c}=\pm\mathrm{1} \\ $$$$\mathrm{for}\:\mathrm{0}\leqslant{x}<\mathrm{2}\pi\:\mathrm{we}\:\mathrm{get} \\ $$$${x}\in\left\{\mathrm{0},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4},\:\mathrm{6},\:\mathrm{8},\:\mathrm{9},\:\mathrm{10}\right\}×\frac{\pi}{\mathrm{6}} \\ $$

Commented by MathSh last updated on 28/Oct/21

$$\mathrm{Perfect}\:\mathrm{dear}\:\boldsymbol{\mathrm{Ser}},\:\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much} \\ $$

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