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Question Number 228155 by hardmath last updated on 31/Mar/26
sinx + cosx = tanx find: x = ?
$$\mathrm{sin}\boldsymbol{\mathrm{x}}\:+\:\mathrm{cos}\boldsymbol{\mathrm{x}}\:=\:\mathrm{tan}\boldsymbol{\mathrm{x}} \\ $$$$\mathrm{find}:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$
Commented by TonyCWX last updated on 01/Apr/26
I will try Weierstrass Substitution. Will update later.
$$\mathrm{I}\:\mathrm{will}\:\mathrm{try}\:\mathrm{Weierstrass}\:\mathrm{Substitution}. \\ $$$$\mathrm{Will}\:\mathrm{update}\:\mathrm{later}. \\ $$
Commented by Abod last updated on 01/Apr/26
hello
$${hello} \\ $$
Commented by Jyrgen last updated on 01/Apr/26
With t=tan (x/2) we end up here: t^4 −4t^3 −2t^2 +1=0 t=u+1 u^4 −8u^2 −12u−4=0 this has useable solutions if v^3 −((16)/3)v−((560)/(27))=0 has at least one useable solution but v=(((280)/(27))+((8(√(129)))/9))^(1/3) +(((280)/(27))−((8(√(129)))/9))^(1/3) ⇒ no useable exact solution exists
$${With}\:{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:{we}\:{end}\:{up}\:{here}: \\ $$$${t}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{3}} −\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$${t}={u}+\mathrm{1} \\ $$$${u}^{\mathrm{4}} −\mathrm{8}{u}^{\mathrm{2}} −\mathrm{12}{u}−\mathrm{4}=\mathrm{0} \\ $$$${this}\:{has}\:{useable}\:{solutions}\:{if} \\ $$$${v}^{\mathrm{3}} −\frac{\mathrm{16}}{\mathrm{3}}{v}−\frac{\mathrm{560}}{\mathrm{27}}=\mathrm{0} \\ $$$${has}\:{at}\:{least}\:{one}\:{useable}\:{solution}\:{but} \\ $$$${v}=\left(\frac{\mathrm{280}}{\mathrm{27}}+\frac{\mathrm{8}\sqrt{\mathrm{129}}}{\mathrm{9}}\right)^{\mathrm{1}/\mathrm{3}} +\left(\frac{\mathrm{280}}{\mathrm{27}}−\frac{\mathrm{8}\sqrt{\mathrm{129}}}{\mathrm{9}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\Rightarrow\:{no}\:{useable}\:{exact}\:{solution}\:{exists} \\ $$
Answered by Jyrgen last updated on 01/Apr/26
sin x +cos x =tan x ((1+tan x)/( (√(1+tan^2 x))))=tan x This leads to tan^4 x −2tan x −1=0 which gives no useable exact solution. Approximating I get x≈0.302066+2nπ x≈0.858943+2nπ with n∈Z
$$\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}\:=\mathrm{tan}\:{x} \\ $$$$\frac{\mathrm{1}+\mathrm{tan}\:{x}}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{x}}}=\mathrm{tan}\:{x} \\ $$$${This}\:{leads}\:{to} \\ $$$$\mathrm{tan}^{\mathrm{4}} \:{x}\:−\mathrm{2tan}\:{x}\:−\mathrm{1}=\mathrm{0} \\ $$$${which}\:{gives}\:{no}\:{useable}\:{exact}\:{solution}. \\ $$$${Approximating}\:{I}\:{get} \\ $$$${x}\approx\mathrm{0}.\mathrm{302066}+\mathrm{2}{n}\pi \\ $$$${x}\approx\mathrm{0}.\mathrm{858943}+\mathrm{2}{n}\pi \\ $$$${with}\:{n}\in\mathbb{Z} \\ $$
Answered by TonyCWX last updated on 01/Apr/26
sin(x) + cos(x) = tan(x) determinant (((t = tan((x/2)))))((2t)/(1 + t^2 )) + ((1 − t^2 )/(1 + t^2 )) = ((2t)/(1 − t^2 )) ⇒ t^4 − 4t^3 − 2t^2 + 1 = 0 determinant (((t = u + 1)))u^4 − 8u^2 − 12u − 4 = 0 By letting u^4 − 8u^2 − 12u − 4 = (u^2 + αu + β)(u^2 + γu + δ), ⇒ { ((α + γ = 0)),((β + δ + αγ = −8)),((αδ + βγ = −12)),((βδ = −4)) :} ⇒ { ((γ = −α)),((δ = ((8α)/(8α − α^3 − 12)))),((α^6 − 16α^4 + 80α^2 − 144 = 0)),((β = ((α^3 − 8α + 12)/(2α)))) :} Now solving: α^6 − 16α^4 + 160α^2 − 144 = 0 determinant (((λ = α^2 )))λ^3 − 16λ^2 + 80λ − 144 = 0 determinant (((λ = ξ + ((16)/3))))ξ^3 − ((16)/3)ξ − ((560)/(27)) = 0 ξ = (( ((−(−((560)/(27)))+(√((−((560)/(27)))^2 + ((4(−((16)/3))^3 )/(27)))))/2)))^(1/3) + (((−(−((560)/(27))) − (√((−((560)/(27)))^2 + ((4(−((16)/3))^3 )/(27)))))/2))^(1/3) ξ = (((4)^(1/3) (((70 − 6(√(129))))^(1/3) + ((70 + 6(√(129))))^(1/3) ))/3) λ = (((4)^(1/3) (((70 − 6(√(129))))^(1/3) + ((70 + 6(√(129))))^(1/3) ) + 16)/3) α = ±(√((((280 − 24(√(129))))^(1/3) + ((280 + 24(√(129))))^(1/3) + 16)/3)) (u^2 + αu + β)(u^2 + γu + δ) = 0 ⇒ u = ((−a ± (√(α^2 − 4β)))/2) ∨ ((−γ ± (√(γ^2 − 4δ)))/2) ⇒ u = ((−a ± (√(α^2 − 2α^3 + 16α − 24)))/2) ∨ ((α ± (√(α^2 − ((32α)/(8α − α^3 − 12)))))/2) ⇒ t = ((2 − a ± (√(α^2 − 2α^3 + 16α − 24)))/2) ∨ ((2 + α ± (√(α^2 − ((32α)/(8α − α^3 − 12)))))/2) ⇒ x = 2(tan^(−1) (((2 − a ± (√(α^2 − 2α^3 + 16α − 24)))/2)) + nπ) ∨ 2(tan^(−1) (((2 + α ± (√(α^2 − ((32α)/(8α − α^3 − 12)))))/2)) + nπ), n ∈ Z
$$\mathrm{sin}\left({x}\right)\:+\:\mathrm{cos}\left({x}\right)\:=\:\mathrm{tan}\left({x}\right) \\ $$$$\begin{array}{|c|}{{t}\:=\:\mathrm{tan}\left(\frac{{x}}{\mathrm{2}}\right)}\\\hline\end{array}\frac{\mathrm{2}{t}}{\mathrm{1}\:+\:{t}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}\:−\:{t}^{\mathrm{2}} }{\mathrm{1}\:+\:{t}^{\mathrm{2}} }\:=\:\frac{\mathrm{2}{t}}{\mathrm{1}\:−\:{t}^{\mathrm{2}} } \\ $$$$\Rightarrow\:{t}^{\mathrm{4}} \:−\:\mathrm{4}{t}^{\mathrm{3}} \:−\:\mathrm{2}{t}^{\mathrm{2}} \:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\begin{array}{|c|}{{t}\:=\:{u}\:+\:\mathrm{1}}\\\hline\end{array}{u}^{\mathrm{4}} \:−\:\mathrm{8}{u}^{\mathrm{2}} \:−\:\mathrm{12}{u}\:−\:\mathrm{4}\:=\:\mathrm{0} \\ $$$$\mathrm{By}\:\mathrm{letting}\:{u}^{\mathrm{4}} \:−\:\mathrm{8}{u}^{\mathrm{2}} \:−\:\mathrm{12}{u}\:−\:\mathrm{4}\:=\:\left({u}^{\mathrm{2}} \:+\:\alpha{u}\:+\:\beta\right)\left({u}^{\mathrm{2}} \:+\:\gamma{u}\:+\:\delta\right), \\ $$$$\Rightarrow\:\begin{cases}{\alpha\:+\:\gamma\:=\:\mathrm{0}}\\{\beta\:+\:\delta\:+\:\alpha\gamma\:=\:−\mathrm{8}}\\{\alpha\delta\:+\:\beta\gamma\:=\:−\mathrm{12}}\\{\beta\delta\:=\:−\mathrm{4}}\end{cases}\:\Rightarrow\:\begin{cases}{\gamma\:=\:−\alpha}\\{\delta\:=\:\frac{\mathrm{8}\alpha}{\mathrm{8}\alpha\:−\:\alpha^{\mathrm{3}} \:−\:\mathrm{12}}}\\{\alpha^{\mathrm{6}} \:−\:\mathrm{16}\alpha^{\mathrm{4}} \:+\:\mathrm{80}\alpha^{\mathrm{2}} \:−\:\mathrm{144}\:=\:\mathrm{0}}\\{\beta\:=\:\frac{\alpha^{\mathrm{3}} \:−\:\mathrm{8}\alpha\:+\:\mathrm{12}}{\mathrm{2}\alpha}}\end{cases} \\ $$$$ \\ $$$$\mathrm{Now}\:\mathrm{solving}:\:\alpha^{\mathrm{6}} \:−\:\mathrm{16}\alpha^{\mathrm{4}} \:+\:\mathrm{160}\alpha^{\mathrm{2}} \:−\:\mathrm{144}\:=\:\mathrm{0} \\ $$$$\begin{array}{|c|}{\lambda\:=\:\alpha^{\mathrm{2}} }\\\hline\end{array}\lambda^{\mathrm{3}} \:−\:\mathrm{16}\lambda^{\mathrm{2}} \:+\:\mathrm{80}\lambda\:−\:\mathrm{144}\:=\:\mathrm{0} \\ $$$$\begin{array}{|c|}{\lambda\:=\:\xi\:+\:\frac{\mathrm{16}}{\mathrm{3}}}\\\hline\end{array}\xi^{\mathrm{3}} \:−\:\frac{\mathrm{16}}{\mathrm{3}}\xi\:−\:\frac{\mathrm{560}}{\mathrm{27}}\:=\:\mathrm{0} \\ $$$$\xi\:=\:\sqrt[{\mathrm{3}}]{\:\frac{−\left(−\frac{\mathrm{560}}{\mathrm{27}}\right)+\sqrt{\left(−\frac{\mathrm{560}}{\mathrm{27}}\right)^{\mathrm{2}} +\:\frac{\mathrm{4}\left(−\frac{\mathrm{16}}{\mathrm{3}}\right)^{\mathrm{3}} }{\mathrm{27}}}}{\mathrm{2}}}\:+\:\sqrt[{\mathrm{3}}]{\frac{−\left(−\frac{\mathrm{560}}{\mathrm{27}}\right)\:−\:\sqrt{\left(−\frac{\mathrm{560}}{\mathrm{27}}\right)^{\mathrm{2}} \:+\:\frac{\mathrm{4}\left(−\frac{\mathrm{16}}{\mathrm{3}}\right)^{\mathrm{3}} }{\mathrm{27}}}}{\mathrm{2}}} \\ $$$$\xi\:=\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{4}}\left(\sqrt[{\mathrm{3}}]{\mathrm{70}\:−\:\mathrm{6}\sqrt{\mathrm{129}}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{70}\:+\:\mathrm{6}\sqrt{\mathrm{129}}}\right)}{\mathrm{3}} \\ $$$$\lambda\:=\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{4}}\left(\sqrt[{\mathrm{3}}]{\mathrm{70}\:−\:\mathrm{6}\sqrt{\mathrm{129}}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{70}\:+\:\mathrm{6}\sqrt{\mathrm{129}}}\right)\:+\:\mathrm{16}}{\mathrm{3}} \\ $$$$\alpha\:=\:\pm\sqrt{\frac{\sqrt[{\mathrm{3}}]{\mathrm{280}\:−\:\mathrm{24}\sqrt{\mathrm{129}}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{280}\:+\:\mathrm{24}\sqrt{\mathrm{129}}}\:+\:\mathrm{16}}{\mathrm{3}}} \\ $$$$ \\ $$$$\left({u}^{\mathrm{2}} \:+\:\alpha{u}\:+\:\beta\right)\left({u}^{\mathrm{2}} \:+\:\gamma{u}\:+\:\delta\right)\:=\:\mathrm{0} \\ $$$$\Rightarrow\:{u}\:=\:\frac{−{a}\:\pm\:\sqrt{\alpha^{\mathrm{2}} \:−\:\mathrm{4}\beta}}{\mathrm{2}}\:\vee\:\frac{−\gamma\:\pm\:\sqrt{\gamma^{\mathrm{2}} \:−\:\mathrm{4}\delta}}{\mathrm{2}} \\ $$$$\Rightarrow\:{u}\:=\:\frac{−{a}\:\pm\:\sqrt{\alpha^{\mathrm{2}} \:−\:\mathrm{2}\alpha^{\mathrm{3}} \:+\:\mathrm{16}\alpha\:−\:\mathrm{24}}}{\mathrm{2}}\:\vee\:\frac{\alpha\:\pm\:\sqrt{\alpha^{\mathrm{2}} \:−\:\frac{\mathrm{32}\alpha}{\mathrm{8}\alpha\:−\:\alpha^{\mathrm{3}} \:−\:\mathrm{12}}}}{\mathrm{2}} \\ $$$$\Rightarrow\:{t}\:=\:\frac{\mathrm{2}\:−\:{a}\:\pm\:\sqrt{\alpha^{\mathrm{2}} \:−\:\mathrm{2}\alpha^{\mathrm{3}} \:+\:\mathrm{16}\alpha\:−\:\mathrm{24}}}{\mathrm{2}}\:\vee\:\frac{\mathrm{2}\:+\:\alpha\:\pm\:\sqrt{\alpha^{\mathrm{2}} \:−\:\frac{\mathrm{32}\alpha}{\mathrm{8}\alpha\:−\:\alpha^{\mathrm{3}} \:−\:\mathrm{12}}}}{\mathrm{2}} \\ $$$$\Rightarrow\:{x}\:=\:\mathrm{2}\left(\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}\:−\:{a}\:\pm\:\sqrt{\alpha^{\mathrm{2}} \:−\:\mathrm{2}\alpha^{\mathrm{3}} \:+\:\mathrm{16}\alpha\:−\:\mathrm{24}}}{\mathrm{2}}\right)\:+\:{n}\pi\right)\:\vee\:\mathrm{2}\left(\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}\:+\:\alpha\:\pm\:\sqrt{\alpha^{\mathrm{2}} \:−\:\frac{\mathrm{32}\alpha}{\mathrm{8}\alpha\:−\:\alpha^{\mathrm{3}} \:−\:\mathrm{12}}}}{\mathrm{2}}\right)\:+\:{n}\pi\right),\:{n}\:\in\:\mathbb{Z} \\ $$
Commented by Jyrgen last updated on 01/Apr/26
But we cannot use these or can we?
$${But}\:{we}\:{cannot}\:{use}\:{these}\:{or}\:{can}\:{we}? \\ $$
Commented by TonyCWX last updated on 01/Apr/26
The first part will provide 2 Complex Solutions, while the second part will provide 2 Real Solutions, which satisfy the values after I used a calculator.
$$\mathrm{The}\:\mathrm{first}\:\mathrm{part}\:\mathrm{will}\:\mathrm{provide}\:\mathrm{2}\:\mathrm{Complex}\:\mathrm{Solutions},\:\mathrm{while}\:\mathrm{the}\:\mathrm{second}\:\mathrm{part}\:\mathrm{will}\:\mathrm{provide}\:\mathrm{2}\:\mathrm{Real}\:\mathrm{Solutions},\:\mathrm{which}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{values}\:\mathrm{after}\:\mathrm{I}\:\mathrm{used}\:\mathrm{a}\:\mathrm{calculator}. \\ $$
Commented by TonyCWX last updated on 01/Apr/26
Commented by TonyCWX last updated on 01/Apr/26
Checks out. My solution is indeed correct.
$$\mathrm{Checks}\:\mathrm{out}. \\ $$$$\mathrm{My}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{indeed}\:\mathrm{correct}. \\ $$
Commented by Jyrgen last updated on 01/Apr/26
Correct: yes Useable: no
$${Correct}:\:{yes} \\ $$$${Useable}:\:{no} \\ $$
Answered by ajfour last updated on 02/Apr/26
s+c=t let s−c=p t=((t+p)/(t−p)) ⇒ p=t(((t−1)/(t+1))) and t^2 +p^2 =2 so t^2 (t^2 +1)=(t+1)^2 t^4 −2t−1=0 (t^2 +st+h)(t^2 −st+k)=0 h+k=s^2 s(h−k)=2 h=(s^2 /2)+(1/s) k=(s^2 /2)−(1/s) hk=−1 s^4 −(4/s^2 )+4=0 s^2 =z z^3 +4z−4=0 z=((√(4+((64)/(27))))+2)^(1/3) −((√(4+((64)/(27))))−2)^(1/3) z=(((2(√(43)))/(3(√3)))+2)^(1/3) −(((2(√(43)))/(3(√3)))−2)^(1/3) s=z^(1/2) t^2 −st+k=0 t=tan x=(s/2)±(√((1/s)−(s^2 /4)))
$${s}+{c}={t} \\ $$$${let}\:{s}−{c}={p} \\ $$$${t}=\frac{{t}+{p}}{{t}−{p}} \\ $$$$\Rightarrow\:{p}={t}\left(\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}\right) \\ $$$${and}\:{t}^{\mathrm{2}} +{p}^{\mathrm{2}} =\mathrm{2} \\ $$$${so} \\ $$$${t}^{\mathrm{2}} \left({t}^{\mathrm{2}} +\mathrm{1}\right)=\left({t}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${t}^{\mathrm{4}} −\mathrm{2}{t}−\mathrm{1}=\mathrm{0} \\ $$$$\left({t}^{\mathrm{2}} +{st}+{h}\right)\left({t}^{\mathrm{2}} −{st}+{k}\right)=\mathrm{0} \\ $$$${h}+{k}={s}^{\mathrm{2}} \\ $$$${s}\left({h}−{k}\right)=\mathrm{2} \\ $$$${h}=\frac{{s}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{1}}{{s}} \\ $$$${k}=\frac{{s}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{1}}{{s}} \\ $$$${hk}=−\mathrm{1} \\ $$$${s}^{\mathrm{4}} −\frac{\mathrm{4}}{{s}^{\mathrm{2}} }+\mathrm{4}=\mathrm{0} \\ $$$${s}^{\mathrm{2}} ={z} \\ $$$${z}^{\mathrm{3}} +\mathrm{4}{z}−\mathrm{4}=\mathrm{0} \\ $$$${z}=\left(\sqrt{\mathrm{4}+\frac{\mathrm{64}}{\mathrm{27}}}+\mathrm{2}\right)^{\mathrm{1}/\mathrm{3}} −\left(\sqrt{\mathrm{4}+\frac{\mathrm{64}}{\mathrm{27}}}−\mathrm{2}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$${z}=\left(\frac{\mathrm{2}\sqrt{\mathrm{43}}}{\mathrm{3}\sqrt{\mathrm{3}}}+\mathrm{2}\right)^{\mathrm{1}/\mathrm{3}} −\left(\frac{\mathrm{2}\sqrt{\mathrm{43}}}{\mathrm{3}\sqrt{\mathrm{3}}}−\mathrm{2}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$${s}={z}^{\mathrm{1}/\mathrm{2}} \\ $$$${t}^{\mathrm{2}} −{st}+{k}=\mathrm{0} \\ $$$${t}=\mathrm{tan}\:{x}=\frac{{s}}{\mathrm{2}}\pm\sqrt{\frac{\mathrm{1}}{{s}}−\frac{{s}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by TonyCWX last updated on 03/Apr/26
Great work, sir! By the way, are you still posting videos on YouTube? It seems like I can′t access to your older videos...
$$\mathrm{Great}\:\mathrm{work},\:\mathrm{sir}! \\ $$$$\mathrm{By}\:\mathrm{the}\:\mathrm{way},\:\mathrm{are}\:\mathrm{you}\:\mathrm{still}\:\mathrm{posting}\:\mathrm{videos}\:\mathrm{on}\:\mathrm{YouTube}? \\ $$$$\mathrm{It}\:\mathrm{seems}\:\mathrm{like}\:\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{access}\:\mathrm{to}\:\mathrm{your}\:\mathrm{older}\:\mathrm{videos}… \\ $$
Commented by ajfour last updated on 03/Apr/26
yeah i deleted those but i just might make more!
$${yeah}\:{i}\:{deleted}\:{those}\:{but}\:{i}\:{just}\:{might} \\ $$$${make}\:{more}! \\ $$
Commented by TonyCWX last updated on 03/Apr/26
I see. Looking forward!
$$\mathrm{I}\:\mathrm{see}. \\ $$$$\mathrm{Looking}\:\mathrm{forward}! \\ $$

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