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Question Number 136008 by liberty last updated on 17/Mar/21

cos x (√(tan x)) = sin^3 x+cos^3 x

$$\:\mathrm{cos}\:{x}\:\sqrt{\mathrm{tan}\:{x}}\:=\:\mathrm{sin}\:^{\mathrm{3}} {x}+\mathrm{cos}\:^{\mathrm{3}} {x} \\ $$

Answered by MJS_new last updated on 17/Mar/21

t=(√(tan x)) ∧t≥0 (t/((t^4 +1)^(1/2) ))=((t^6 +1)/((t^4 +1)^(3/2) )) t^6 −t^5 −t+1=0 (t−1)(t^5 −1)=0 ⇒ t=1 ⇒ x=(π/4)+nπ

$${t}=\sqrt{\mathrm{tan}\:{x}}\:\wedge{t}\geqslant\mathrm{0} \\ $$$$\frac{{t}}{\left({t}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{1}/\mathrm{2}} }=\frac{{t}^{\mathrm{6}} +\mathrm{1}}{\left({t}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} } \\ $$$${t}^{\mathrm{6}} −{t}^{\mathrm{5}} −{t}+\mathrm{1}=\mathrm{0} \\ $$$$\left({t}−\mathrm{1}\right)\left({t}^{\mathrm{5}} −\mathrm{1}\right)=\mathrm{0}\:\Rightarrow\:{t}=\mathrm{1} \\ $$$$\Rightarrow\:{x}=\frac{\pi}{\mathrm{4}}+{n}\pi \\ $$

Answered by liberty last updated on 18/Mar/21

note that tan x ≥ 0 ⇒ (√(sin x cos x)) = (sin x+cos x)(1−sin x cos x) ⇒(√(sin x cos x)) =(√(1+2sin x cos x)) (1−sin x cos x ) let sin x cos x = ℓ ⇒(√ℓ) = (√(1+2ℓ)) (1−ℓ) ⇒ℓ = (1+2ℓ)(1−2ℓ+ℓ^2 ) ⇒ℓ = 1−2ℓ+ℓ^2 +2ℓ−4ℓ^2 +2ℓ^3 ⇒2ℓ^3 −3ℓ^2 −ℓ+1=0 ⇒(2ℓ−1)(ℓ^2 −ℓ−1)=0 ⇒ { ((ℓ=(1/2)⇒sin xcos x=(1/2), sin 2x=1)),((ℓ = ((1±(√5))/2) (rejected))) :} so solution is 2x = (π/2)+2nπ or x = (π/4)+nπ

$${note}\:{that}\:\mathrm{tan}\:{x}\:\geqslant\:\mathrm{0} \\ $$$$\Rightarrow\:\sqrt{\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}}\:=\:\left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right)\left(\mathrm{1}−\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}\right) \\ $$$$\Rightarrow\sqrt{\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}}\:=\sqrt{\mathrm{1}+\mathrm{2sin}\:{x}\:\mathrm{cos}\:{x}}\:\left(\mathrm{1}−\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}\:\right) \\ $$$${let}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}\:=\:\ell \\ $$$$\Rightarrow\sqrt{\ell}\:=\:\sqrt{\mathrm{1}+\mathrm{2}\ell}\:\left(\mathrm{1}−\ell\right) \\ $$$$\Rightarrow\ell\:=\:\left(\mathrm{1}+\mathrm{2}\ell\right)\left(\mathrm{1}−\mathrm{2}\ell+\ell^{\mathrm{2}} \right) \\ $$$$\Rightarrow\ell\:=\:\mathrm{1}−\mathrm{2}\ell+\ell^{\mathrm{2}} +\mathrm{2}\ell−\mathrm{4}\ell^{\mathrm{2}} +\mathrm{2}\ell^{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{2}\ell^{\mathrm{3}} −\mathrm{3}\ell^{\mathrm{2}} −\ell+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{2}\ell−\mathrm{1}\right)\left(\ell^{\mathrm{2}} −\ell−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\begin{cases}{\ell=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{sin}\:{x}\mathrm{cos}\:{x}=\frac{\mathrm{1}}{\mathrm{2}},\:\mathrm{sin}\:\mathrm{2}{x}=\mathrm{1}}\\{\ell\:=\:\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}\:\left({rejected}\right)}\end{cases} \\ $$$${so}\:{solution}\:{is}\:\mathrm{2}{x}\:=\:\frac{\pi}{\mathrm{2}}+\mathrm{2}{n}\pi\: \\ $$$${or}\:{x}\:=\:\frac{\pi}{\mathrm{4}}+{n}\pi \\ $$

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