Question-203897

Question Number 203897 by NasaSara last updated on 01/Feb/24

Answered by AST last updated on 01/Feb/24

cos^2 (x)−cos(x)=(√(1−cos^2 x)) ⇒cos^4 (x)+2cos^2 (x)−2cos^3 (x)−1=0 For x∈Z⇒cos(x)=1⇒x=0+2nπ(n∈Z)

$${cos}^{\mathrm{2}} \left({x}\right)−{cos}\left({x}\right)=\sqrt{\mathrm{1}−{cos}^{\mathrm{2}} {x}} \\ $$$$\Rightarrow{cos}^{\mathrm{4}} \left({x}\right)+\mathrm{2}{cos}^{\mathrm{2}} \left({x}\right)−\mathrm{2}{cos}^{\mathrm{3}} \left({x}\right)−\mathrm{1}=\mathrm{0} \\ $$$${For}\:{x}\in\mathbb{Z}\Rightarrow{cos}\left({x}\right)=\mathrm{1}\Rightarrow{x}=\mathrm{0}+\mathrm{2}{n}\pi\left({n}\in\mathbb{Z}\right) \\ $$

Commented by NasaSara last updated on 01/Feb/24

thx

Answered by Frix last updated on 01/Feb/24

Obviously x=2nπ Use t=tan (x/2) ⇒ t(t^3 −t^2 −t−1)=0 t=0 ⇒ x=2nπ t=(1/3)+(((29)/(27))+((√(33))/9))^(1/3) +(((29)/(27))−((√(33))/9))^(1/3) t≈1.83929 ⇒ x≈2.14562+2nπ

$$\mathrm{Obviously}\:{x}=\mathrm{2}{n}\pi \\ $$$$ \\ $$$$\mathrm{Use}\:{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\Rightarrow \\ $$$${t}\left({t}^{\mathrm{3}} −{t}^{\mathrm{2}} −{t}−\mathrm{1}\right)=\mathrm{0} \\ $$$${t}=\mathrm{0}\:\Rightarrow\:{x}=\mathrm{2}{n}\pi \\ $$$${t}=\frac{\mathrm{1}}{\mathrm{3}}+\left(\frac{\mathrm{29}}{\mathrm{27}}+\frac{\sqrt{\mathrm{33}}}{\mathrm{9}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} +\left(\frac{\mathrm{29}}{\mathrm{27}}−\frac{\sqrt{\mathrm{33}}}{\mathrm{9}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$${t}\approx\mathrm{1}.\mathrm{83929}\:\Rightarrow\:{x}\approx\mathrm{2}.\mathrm{14562}+\mathrm{2}{n}\pi \\ $$

Answered by Frix last updated on 01/Feb/24

Another weird path came to mind... cos x =u ∧sin x =(√(1−u^2 ))=(√(1−u))(√(1+u)) tan (x/2) =((√(1−u))/( (√(1+u)))) ((cos^2 x −cos x −sin x)/(tan (x/2)))=((u^2 −u−(√(1−u))(√(1+u)))/((√(1−u))/( (√(1+u)))))= =−(u+1+u(√(1−u))(√(1+u))) cos^2 x −cos x −sin x=0 −(1+(1+sin x)cos x )tan (x/2) =0 tan (x/2) =0 ⇒ x=2nπ u+1=−u(√(1−u))(√(1+u)) (u+1)^2 =u^2 (1−u)(1+u) u+1=u^2 (1−u) u^3 −u^2 +u+1=0 u≈−.543689=cos x x≈2.14562+2nπ

$$\mathrm{Another}\:\mathrm{weird}\:\mathrm{path}\:\mathrm{came}\:\mathrm{to}\:\mathrm{mind}… \\ $$$$\mathrm{cos}\:{x}\:={u}\:\wedge\mathrm{sin}\:{x}\:=\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }=\sqrt{\mathrm{1}−{u}}\sqrt{\mathrm{1}+{u}} \\ $$$$\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:=\frac{\sqrt{\mathrm{1}−{u}}}{\:\sqrt{\mathrm{1}+{u}}} \\ $$$$\frac{\mathrm{cos}^{\mathrm{2}} \:{x}\:−\mathrm{cos}\:{x}\:−\mathrm{sin}\:{x}}{\mathrm{tan}\:\frac{{x}}{\mathrm{2}}}=\frac{{u}^{\mathrm{2}} −{u}−\sqrt{\mathrm{1}−{u}}\sqrt{\mathrm{1}+{u}}}{\frac{\sqrt{\mathrm{1}−{u}}}{\:\sqrt{\mathrm{1}+{u}}}}= \\ $$$$=−\left({u}+\mathrm{1}+{u}\sqrt{\mathrm{1}−{u}}\sqrt{\mathrm{1}+{u}}\right) \\ $$$$ \\ $$$$\mathrm{cos}^{\mathrm{2}} \:{x}\:−\mathrm{cos}\:{x}\:−\mathrm{sin}\:{x}=\mathrm{0} \\ $$$$−\left(\mathrm{1}+\left(\mathrm{1}+\mathrm{sin}\:{x}\right)\mathrm{cos}\:{x}\:\right)\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:=\mathrm{0} \\ $$$$\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:=\mathrm{0}\:\Rightarrow\:{x}=\mathrm{2}{n}\pi \\ $$$$ \\ $$$${u}+\mathrm{1}=−{u}\sqrt{\mathrm{1}−{u}}\sqrt{\mathrm{1}+{u}} \\ $$$$\left({u}+\mathrm{1}\right)^{\mathrm{2}} ={u}^{\mathrm{2}} \left(\mathrm{1}−{u}\right)\left(\mathrm{1}+{u}\right) \\ $$$${u}+\mathrm{1}={u}^{\mathrm{2}} \left(\mathrm{1}−{u}\right) \\ $$$${u}^{\mathrm{3}} −{u}^{\mathrm{2}} +{u}+\mathrm{1}=\mathrm{0} \\ $$$${u}\approx−.\mathrm{543689}=\mathrm{cos}\:{x} \\ $$$${x}\approx\mathrm{2}.\mathrm{14562}+\mathrm{2}{n}\pi \\ $$

Commented by NasaSara last updated on 01/Feb/24

when i put the equation in wolfram alpha some of the solutions where complex ,so how to get the complex solutions ?

Commented by Frix last updated on 02/Feb/24

$$\mathrm{Solve} \\ $$$${t}^{\mathrm{3}} −{t}^{\mathrm{2}} −{t}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{for}\:{t}\in\mathbb{C} \\ $$

Commented by NasaSara last updated on 02/Feb/24

thank you

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