Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 157489 by MathSh last updated on 23/Oct/21

Solve for real numbers:  tan(10x) = tan^5 (2x)

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\mathrm{tan}\left(\mathrm{10x}\right)\:=\:\mathrm{tan}^{\mathrm{5}} \left(\mathrm{2x}\right) \\ $$$$ \\ $$

Answered by MJS_new last updated on 24/Oct/21

let t=tan 2x  ((t(t^4 −10t^2 +5))/(5t^4 −10t^2 +1))=t^5   (t+1)^3 t(t−1)^3 (t^2 +1)=0  t=−1∨t=0∨t=1  x=−(π/8)+((nπ)/2)∨x=((nπ)/2)∨x=(π/8)+((nπ)/2)

$$\mathrm{let}\:{t}=\mathrm{tan}\:\mathrm{2}{x} \\ $$$$\frac{{t}\left({t}^{\mathrm{4}} −\mathrm{10}{t}^{\mathrm{2}} +\mathrm{5}\right)}{\mathrm{5}{t}^{\mathrm{4}} −\mathrm{10}{t}^{\mathrm{2}} +\mathrm{1}}={t}^{\mathrm{5}} \\ $$$$\left({t}+\mathrm{1}\right)^{\mathrm{3}} {t}\left({t}−\mathrm{1}\right)^{\mathrm{3}} \left({t}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{0} \\ $$$${t}=−\mathrm{1}\vee{t}=\mathrm{0}\vee{t}=\mathrm{1} \\ $$$${x}=−\frac{\pi}{\mathrm{8}}+\frac{{n}\pi}{\mathrm{2}}\vee{x}=\frac{{n}\pi}{\mathrm{2}}\vee{x}=\frac{\pi}{\mathrm{8}}+\frac{{n}\pi}{\mathrm{2}} \\ $$

Commented by tounghoungko last updated on 24/Oct/21

tan 5x=((tan^5 x−10tan^3 x+5tan x)/(1−10tan^2 x+5tan^4 x))

$$\mathrm{tan}\:\mathrm{5}{x}=\frac{\mathrm{tan}\:^{\mathrm{5}} {x}−\mathrm{10tan}\:^{\mathrm{3}} {x}+\mathrm{5tan}\:{x}}{\mathrm{1}−\mathrm{10tan}\:^{\mathrm{2}} {x}+\mathrm{5tan}\:^{\mathrm{4}} {x}} \\ $$

Commented by MJS_new last updated on 24/Oct/21

yes

$$\mathrm{yes} \\ $$

Commented by MathSh last updated on 24/Oct/21

Thank you dear Ser very nice

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er}\:\mathrm{very}\:\mathrm{nice} \\ $$

Answered by tounghoungko last updated on 24/Oct/21

 Solve for x∈R . tan (10x)=tan^5 (2x)  ⇔ tan (5θ)=((sin^5 θ)/(cos^5 θ)) ,[θ=2x]   ⇔((sin 5θ)/(cos 5θ)) = ((sin^5 θ)/(cos^5 θ))  ⇔ ((5sin θ−20sin^3 θ+16sin^5 θ)/(5cos θ−20cos^3 θ+16cos^5 θ))=((sin^5 θ)/(cos^5 θ))  ⇔5sin θcos^5 θ−20sin^3 θcos^5 θ+16sin^5 θcos^5 θ=       5cos θsin^5 θ−20cos^3 θsin^5 θ+16cos^5 θsin^5 θ  ⇔sin θcos θ [5cos^4 θ−20sin^2 θcos^4 θ+16sin^4 θcos^4 θ−                            [5sin^4 θ+20cos^2 θsin^4 θ−16sin^4 θcos^4 θ ]=0  ⇔ (1/2)sin 2θ [5(cos^4 θ−sin^4 θ)+20(cos^2 θsin^4 θ−sin^2 θcos^4 θ)]=0  ⇔(1/2)sin 2θ [5(cos^2 θ−sin^2 θ)+20cos^2 θsin^2 θ(sin^2 θ−cos^2 θ)]=0  ⇔(5/2)sin 2θ cos 2θ (1−sin^2 2θ)=0  ⇔(5/4)sin 4θ (1−sin^2 2θ)=0  (i) (5/4)sin 4θ=0⇒  { ((4θ=2nπ)),((4θ=(1+2n)π)) :}   ⇒  { ((8x=2nπ)),((8x=(1+2n)π)) :}⇒ { ((x=((nπ)/4))),((x=(((1+2n)/8))π)) :}  (ii) 1−sin^2 2θ=0  ⇒ { ((sin 2θ=1)),((sin 2θ=−1)) :} ⇒ { (( 2θ=(π/2)+2kπ)),((2θ=((3π)/2)+2kπ)) :}    { ((4x=(π/2)+2kπ)),((4x=((3π)/2)+2kπ)) :}⇒ { ((x=(((4k+1)/8))π)),((x=(((3+4k)/8))π)) :}

$$\:{Solve}\:{for}\:{x}\in{R}\:.\:\mathrm{tan}\:\left(\mathrm{10}{x}\right)=\mathrm{tan}\:^{\mathrm{5}} \left(\mathrm{2}{x}\right) \\ $$$$\Leftrightarrow\:\mathrm{tan}\:\left(\mathrm{5}\theta\right)=\frac{\mathrm{sin}\:^{\mathrm{5}} \theta}{\mathrm{cos}\:^{\mathrm{5}} \theta}\:,\left[\theta=\mathrm{2}{x}\right]\: \\ $$$$\Leftrightarrow\frac{\mathrm{sin}\:\mathrm{5}\theta}{\mathrm{cos}\:\mathrm{5}\theta}\:=\:\frac{\mathrm{sin}\:^{\mathrm{5}} \theta}{\mathrm{cos}\:^{\mathrm{5}} \theta} \\ $$$$\Leftrightarrow\:\frac{\mathrm{5sin}\:\theta−\mathrm{20sin}\:^{\mathrm{3}} \theta+\mathrm{16sin}\:^{\mathrm{5}} \theta}{\mathrm{5cos}\:\theta−\mathrm{20cos}\:^{\mathrm{3}} \theta+\mathrm{16cos}\:^{\mathrm{5}} \theta}=\frac{\mathrm{sin}\:^{\mathrm{5}} \theta}{\mathrm{cos}\:^{\mathrm{5}} \theta} \\ $$$$\Leftrightarrow\mathrm{5sin}\:\theta\mathrm{cos}\:^{\mathrm{5}} \theta−\mathrm{20sin}\:^{\mathrm{3}} \theta\mathrm{cos}\:^{\mathrm{5}} \theta+\mathrm{16sin}\:^{\mathrm{5}} \theta\mathrm{cos}\:^{\mathrm{5}} \theta= \\ $$$$\:\:\:\:\:\mathrm{5cos}\:\theta\mathrm{sin}\:^{\mathrm{5}} \theta−\mathrm{20cos}\:^{\mathrm{3}} \theta\mathrm{sin}\:^{\mathrm{5}} \theta+\mathrm{16cos}\:^{\mathrm{5}} \theta\mathrm{sin}\:^{\mathrm{5}} \theta \\ $$$$\Leftrightarrow\mathrm{sin}\:\theta\mathrm{cos}\:\theta\:\left[\mathrm{5cos}\:^{\mathrm{4}} \theta−\mathrm{20sin}\:^{\mathrm{2}} \theta\mathrm{cos}\:^{\mathrm{4}} \theta+\mathrm{16sin}\:^{\mathrm{4}} \theta\mathrm{cos}\:^{\mathrm{4}} \theta−\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\mathrm{5sin}\:^{\mathrm{4}} \theta+\mathrm{20cos}\:^{\mathrm{2}} \theta\mathrm{sin}\:^{\mathrm{4}} \theta−\mathrm{16sin}\:^{\mathrm{4}} \theta\mathrm{cos}\:^{\mathrm{4}} \theta\:\right]=\mathrm{0} \\ $$$$\Leftrightarrow\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}\theta\:\left[\mathrm{5}\left(\mathrm{cos}\:^{\mathrm{4}} \theta−\mathrm{sin}\:^{\mathrm{4}} \theta\right)+\mathrm{20}\left(\mathrm{cos}^{\mathrm{2}} \theta\mathrm{sin}\:^{\mathrm{4}} \theta−\mathrm{sin}\:^{\mathrm{2}} \theta\mathrm{cos}\:^{\mathrm{4}} \theta\right)\right]=\mathrm{0} \\ $$$$\Leftrightarrow\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}\theta\:\left[\mathrm{5}\left(\mathrm{cos}\:^{\mathrm{2}} \theta−\mathrm{sin}\:^{\mathrm{2}} \theta\right)+\mathrm{20cos}\:^{\mathrm{2}} \theta\mathrm{sin}\:^{\mathrm{2}} \theta\left(\mathrm{sin}\:^{\mathrm{2}} \theta−\mathrm{cos}\:^{\mathrm{2}} \theta\right)\right]=\mathrm{0} \\ $$$$\Leftrightarrow\frac{\mathrm{5}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}\theta\:\mathrm{cos}\:\mathrm{2}\theta\:\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}\theta\right)=\mathrm{0} \\ $$$$\Leftrightarrow\frac{\mathrm{5}}{\mathrm{4}}\mathrm{sin}\:\mathrm{4}\theta\:\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}\theta\right)=\mathrm{0} \\ $$$$\left({i}\right)\:\frac{\mathrm{5}}{\mathrm{4}}\mathrm{sin}\:\mathrm{4}\theta=\mathrm{0}\Rightarrow\:\begin{cases}{\mathrm{4}\theta=\mathrm{2}{n}\pi}\\{\mathrm{4}\theta=\left(\mathrm{1}+\mathrm{2}{n}\right)\pi}\end{cases}\: \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{8}{x}=\mathrm{2}{n}\pi}\\{\mathrm{8}{x}=\left(\mathrm{1}+\mathrm{2}{n}\right)\pi}\end{cases}\Rightarrow\begin{cases}{{x}=\frac{{n}\pi}{\mathrm{4}}}\\{{x}=\left(\frac{\mathrm{1}+\mathrm{2}{n}}{\mathrm{8}}\right)\pi}\end{cases} \\ $$$$\left({ii}\right)\:\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}\theta=\mathrm{0} \\ $$$$\Rightarrow\begin{cases}{\mathrm{sin}\:\mathrm{2}\theta=\mathrm{1}}\\{\mathrm{sin}\:\mathrm{2}\theta=−\mathrm{1}}\end{cases}\:\Rightarrow\begin{cases}{\:\mathrm{2}\theta=\frac{\pi}{\mathrm{2}}+\mathrm{2}{k}\pi}\\{\mathrm{2}\theta=\frac{\mathrm{3}\pi}{\mathrm{2}}+\mathrm{2}{k}\pi}\end{cases} \\ $$$$\:\begin{cases}{\mathrm{4}{x}=\frac{\pi}{\mathrm{2}}+\mathrm{2}{k}\pi}\\{\mathrm{4}{x}=\frac{\mathrm{3}\pi}{\mathrm{2}}+\mathrm{2}{k}\pi}\end{cases}\Rightarrow\begin{cases}{{x}=\left(\frac{\mathrm{4}{k}+\mathrm{1}}{\mathrm{8}}\right)\pi}\\{{x}=\left(\frac{\mathrm{3}+\mathrm{4}{k}}{\mathrm{8}}\right)\pi}\end{cases} \\ $$

Commented by MathSh last updated on 24/Oct/21

Thank you dear Ser cool

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er}\:\mathrm{cool} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com