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Question Number 18962 by Tinkutara last updated on 02/Aug/17
The number of solutions of the equation sin^5 θ + (1/(sin θ)) = (1/(cos θ)) + cos^5 θ where θ ∈ (0, (π/2)) , is
$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{sin}^{\mathrm{5}} \:\theta\:+\:\frac{\mathrm{1}}{\mathrm{sin}\:\theta}\:=\:\frac{\mathrm{1}}{\mathrm{cos}\:\theta}\:+\:\mathrm{cos}^{\mathrm{5}} \:\theta\:\mathrm{where} \\ $$$$\theta\:\in\:\left(\mathrm{0},\:\frac{\pi}{\mathrm{2}}\right)\:,\:\mathrm{is} \\ $$
Answered by behi.8.3.4.1.7@gmail.com last updated on 02/Aug/17
sin^6 θ.cosθ+cosθ=sinθ+cos^6 θ.sinθ ⇒sinθ.cosθ(sin^5 θ−cos^5 θ)−(sinθ−cosθ)=0 sinθ=a,cosθ=b ⇒ab(a^5 −b^5 )−(a−b)=0 ⇒[ab(a^4 +a^3 b+a^2 b^2 +ab^3 +b^4 )−1](a−b)=0 1)a=b⇒sinθ=cosθ⇒tgθ=1⇒θ=kπ+(π/4),k∈Z 2)a^4 +b^4 =(a^2 +b^2 )^2 −2a^2 b^2 =1−2a^2 b^2 a^3 b+ab^3 =ab(a^2 +b^2 )=ab ⇒ab(1−2a^2 b^2 +a^2 b^2 +ab)−1=0 ⇒ab(1+ab−a^2 b^2 )−1=0 (ab=t) ⇒t(1+t−t^2 )−1=0⇒−t^3 +t^2 +t−1=0 ⇒t^3 −t^2 −t+1=0⇒t^2 (t−1)−(t−1)=0 ⇒(t−1)(t^2 −1)=0⇒(t−1)^2 (t+1)=0 ⇒t=1,−1⇒sinθ.cosθ=±1⇒sin2θ=±2 this eq. has not real answers.
$${sin}^{\mathrm{6}} \theta.{cos}\theta+{cos}\theta={sin}\theta+{cos}^{\mathrm{6}} \theta.{sin}\theta \\ $$$$\Rightarrow{sin}\theta.{cos}\theta\left({sin}^{\mathrm{5}} \theta−{cos}^{\mathrm{5}} \theta\right)−\left({sin}\theta−{cos}\theta\right)=\mathrm{0} \\ $$$${sin}\theta={a},{cos}\theta={b} \\ $$$$\Rightarrow{ab}\left({a}^{\mathrm{5}} −{b}^{\mathrm{5}} \right)−\left({a}−{b}\right)=\mathrm{0} \\ $$$$\Rightarrow\left[{ab}\left({a}^{\mathrm{4}} +{a}^{\mathrm{3}} {b}+{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{ab}^{\mathrm{3}} +{b}^{\mathrm{4}} \right)−\mathrm{1}\right]\left({a}−{b}\right)=\mathrm{0} \\ $$$$\left.\mathrm{1}\right){a}={b}\Rightarrow{sin}\theta={cos}\theta\Rightarrow{tg}\theta=\mathrm{1}\Rightarrow\theta={k}\pi+\frac{\pi}{\mathrm{4}},{k}\in{Z} \\ $$$$\left.\mathrm{2}\right){a}^{\mathrm{4}} +{b}^{\mathrm{4}} =\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} =\mathrm{1}−\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$${a}^{\mathrm{3}} {b}+{ab}^{\mathrm{3}} ={ab}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)={ab} \\ $$$$\Rightarrow{ab}\left(\mathrm{1}−\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{ab}\right)−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{ab}\left(\mathrm{1}+{ab}−{a}^{\mathrm{2}} {b}^{\mathrm{2}} \right)−\mathrm{1}=\mathrm{0}\:\:\:\left({ab}={t}\right) \\ $$$$\Rightarrow{t}\left(\mathrm{1}+{t}−{t}^{\mathrm{2}} \right)−\mathrm{1}=\mathrm{0}\Rightarrow−{t}^{\mathrm{3}} +{t}^{\mathrm{2}} +{t}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{t}^{\mathrm{3}} −{t}^{\mathrm{2}} −{t}+\mathrm{1}=\mathrm{0}\Rightarrow{t}^{\mathrm{2}} \left({t}−\mathrm{1}\right)−\left({t}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\left({t}−\mathrm{1}\right)\left({t}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{0}\Rightarrow\left({t}−\mathrm{1}\right)^{\mathrm{2}} \left({t}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{t}=\mathrm{1},−\mathrm{1}\Rightarrow{sin}\theta.{cos}\theta=\pm\mathrm{1}\Rightarrow{sin}\mathrm{2}\theta=\pm\mathrm{2} \\ $$$${this}\:{eq}.\:{has}\:{not}\:{real}\:{answers}. \\ $$
Commented by Tinkutara last updated on 02/Aug/17
Thank you very much behi Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{behi}\:\mathrm{Sir}! \\ $$
Commented by behi.8.3.4.1.7@gmail.com last updated on 04/Aug/17
sin2θ=2⇒((e^(2θ) −e^(−2θ) )/(2i))=2 (e^(2θ) =a) ⇒((a−a^(−1) )/(2i))=2⇒a^2 −4ai−1=0 ⇒a=((4i±(√(−16+4)))/2)=((4i±2(√3)i)/2)=(2±(√3))i ⇒e^(2θ) =(2±(√3))i⇒2θ=ln(2±(√3))+lni ⇒θ=(1/2)ln(2±(√3))+((iπ)/4) (complex answer).
$${sin}\mathrm{2}\theta=\mathrm{2}\Rightarrow\frac{{e}^{\mathrm{2}\theta} −{e}^{−\mathrm{2}\theta} }{\mathrm{2}{i}}=\mathrm{2}\:\:\:\left({e}^{\mathrm{2}\theta} ={a}\right) \\ $$$$\Rightarrow\frac{{a}−{a}^{−\mathrm{1}} }{\mathrm{2}{i}}=\mathrm{2}\Rightarrow{a}^{\mathrm{2}} −\mathrm{4}{ai}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{a}=\frac{\mathrm{4}{i}\pm\sqrt{−\mathrm{16}+\mathrm{4}}}{\mathrm{2}}=\frac{\mathrm{4}{i}\pm\mathrm{2}\sqrt{\mathrm{3}}{i}}{\mathrm{2}}=\left(\mathrm{2}\pm\sqrt{\mathrm{3}}\right){i} \\ $$$$\Rightarrow{e}^{\mathrm{2}\theta} =\left(\mathrm{2}\pm\sqrt{\mathrm{3}}\right){i}\Rightarrow\mathrm{2}\theta={ln}\left(\mathrm{2}\pm\sqrt{\mathrm{3}}\right)+{lni} \\ $$$$\Rightarrow\theta=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\pm\sqrt{\mathrm{3}}\right)+\frac{{i}\pi}{\mathrm{4}}\:\left({complex}\:{answer}\right). \\ $$
Commented by Tinkutara last updated on 04/Aug/17
But using this value in calculator does not gives 2. sin (ln (2 + (√3)) + ((iπ)/2)) = 2.423 + 0.578i
$$\mathrm{But}\:\mathrm{using}\:\mathrm{this}\:\mathrm{value}\:\mathrm{in}\:\mathrm{calculator}\:\mathrm{does} \\ $$$$\mathrm{not}\:\mathrm{gives}\:\mathrm{2}.\:\mathrm{sin}\:\left(\mathrm{ln}\:\left(\mathrm{2}\:+\:\sqrt{\mathrm{3}}\right)\:+\:\frac{{i}\pi}{\mathrm{2}}\right) \\ $$$$=\:\mathrm{2}.\mathrm{423}\:+\:\mathrm{0}.\mathrm{578}{i} \\ $$

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