Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 135679 by liberty last updated on 15/Mar/21

(cos x−sin x)(2tan x+(1/(cos x)))+2 = 0

$$\left(\mathrm{cos}\:{x}−\mathrm{sin}\:{x}\right)\left(\mathrm{2tan}\:{x}+\frac{\mathrm{1}}{\mathrm{cos}\:{x}}\right)+\mathrm{2}\:=\:\mathrm{0} \\ $$

Answered by EDWIN88 last updated on 15/Mar/21

Let tan (x/2) = t → { ((cos x=((1−t)/(1+t^2 )))),((sin x=((2t)/(1+t^2 )))),((tan x=((2t)/(1−t^2 )))) :}  ⇔ (((1−t^2 )/(1+t^2 ))−((2t)/(1+t^2 )))(((4t)/(1−t^2 ))+((1+t^2 )/(1−t)))+2 = 0  ⇒(1−t^2 −2t).(t^2 +4t+1)+2(1−t^4 ) = 0  ⇒3t^4 −t^2 +6t^3 −2t+9t^2 −3 = 0  ⇒t^2 (3t^2 −1)+2t(3t^2 −1)+3(3t^2 −1) = 0  ⇒(3t^2 −1)(t^2 +2t+3) = 0    { ((t^2 +2t+3 = 0 →has complex roots)),((3t^2 −1=0⇒t = ± (1/( (√3))))) :}  ⇔ tan ((x/2)) = ± (1/( (√3))) ; (x/2) = nπ ± (π/6)  ⇒ x = 2nπ ± (π/3) ; n∈Z

$$\mathrm{Let}\:\mathrm{tan}\:\frac{\mathrm{x}}{\mathrm{2}}\:=\:\mathrm{t}\:\rightarrow\begin{cases}{\mathrm{cos}\:\mathrm{x}=\frac{\mathrm{1}−\mathrm{t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }}\\{\mathrm{sin}\:\mathrm{x}=\frac{\mathrm{2t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }}\\{\mathrm{tan}\:\mathrm{x}=\frac{\mathrm{2t}}{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }}\end{cases} \\ $$$$\Leftrightarrow\:\left(\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }−\frac{\mathrm{2t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\right)\left(\frac{\mathrm{4t}}{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }+\frac{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }{\mathrm{1}−\mathrm{t}}\right)+\mathrm{2}\:=\:\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} −\mathrm{2t}\right).\left(\mathrm{t}^{\mathrm{2}} +\mathrm{4t}+\mathrm{1}\right)+\mathrm{2}\left(\mathrm{1}−\mathrm{t}^{\mathrm{4}} \right)\:=\:\mathrm{0} \\ $$$$\Rightarrow\mathrm{3t}^{\mathrm{4}} −\mathrm{t}^{\mathrm{2}} +\mathrm{6t}^{\mathrm{3}} −\mathrm{2t}+\mathrm{9t}^{\mathrm{2}} −\mathrm{3}\:=\:\mathrm{0} \\ $$$$\Rightarrow\mathrm{t}^{\mathrm{2}} \left(\mathrm{3t}^{\mathrm{2}} −\mathrm{1}\right)+\mathrm{2t}\left(\mathrm{3t}^{\mathrm{2}} −\mathrm{1}\right)+\mathrm{3}\left(\mathrm{3t}^{\mathrm{2}} −\mathrm{1}\right)\:=\:\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{3t}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{t}^{\mathrm{2}} +\mathrm{2t}+\mathrm{3}\right)\:=\:\mathrm{0} \\ $$$$\:\begin{cases}{\mathrm{t}^{\mathrm{2}} +\mathrm{2t}+\mathrm{3}\:=\:\mathrm{0}\:\rightarrow\mathrm{has}\:\mathrm{complex}\:\mathrm{roots}}\\{\mathrm{3t}^{\mathrm{2}} −\mathrm{1}=\mathrm{0}\Rightarrow\mathrm{t}\:=\:\pm\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}\end{cases} \\ $$$$\Leftrightarrow\:\mathrm{tan}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\:=\:\pm\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:;\:\frac{\mathrm{x}}{\mathrm{2}}\:=\:\mathrm{n}\pi\:\pm\:\frac{\pi}{\mathrm{6}} \\ $$$$\Rightarrow\:\mathrm{x}\:=\:\mathrm{2n}\pi\:\pm\:\frac{\pi}{\mathrm{3}}\:;\:\mathrm{n}\in\mathbb{Z} \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com