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Question Number 81062 by john santu last updated on 09/Feb/20
cos x (√(1+sin x−2cos x)) = cos x−sin x in [ −5π , −((7π)/2)]
$$\mathrm{cos}\:{x}\:\sqrt{\mathrm{1}+\mathrm{sin}\:{x}−\mathrm{2cos}\:{x}}\:=\:\mathrm{cos}\:{x}−\mathrm{sin}\:{x} \\ $$$${in}\:\left[\:−\mathrm{5}\pi\:,\:−\frac{\mathrm{7}\pi}{\mathrm{2}}\right]\: \\ $$
Commented by MJS last updated on 09/Feb/20
sorry no time today (1) let x=2arctan t ⇒ sin x =((2t)/(t^2 +1)) ∧ cos x =−((t^2 −1)/(t^2 +1)) (2) check for which values the (√(...)) ∈R (3) solve for t [squaring ⇒ not all solutions are useable!] (4) t=tan (x/2) ⇔ x=2nπ+2arctan t find the solution in the required interval
$$\mathrm{sorry}\:\mathrm{no}\:\mathrm{time}\:\mathrm{today} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{let}\:{x}=\mathrm{2arctan}\:{t} \\ $$$$\:\:\:\:\:\Rightarrow\:\mathrm{sin}\:{x}\:=\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{1}}\:\wedge\:\mathrm{cos}\:{x}\:=−\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{check}\:\mathrm{for}\:\mathrm{which}\:\mathrm{values}\:\mathrm{the}\:\sqrt{…}\:\in\mathbb{R} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{solve}\:\mathrm{for}\:{t} \\ $$$$\:\:\:\:\:\left[\mathrm{squaring}\:\Rightarrow\:\mathrm{not}\:\mathrm{all}\:\mathrm{solutions}\:\mathrm{are}\:\mathrm{useable}!\right] \\ $$$$\left(\mathrm{4}\right)\:{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\Leftrightarrow\:{x}=\mathrm{2}{n}\pi+\mathrm{2arctan}\:{t} \\ $$$$\:\:\:\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{in}\:\mathrm{the}\:\mathrm{required}\:\mathrm{interval} \\ $$
Commented by jagoll last updated on 09/Feb/20
hello mister. how are you?
$${hello}\:{mister}.\:{how}\:{are}\:{you}? \\ $$
Commented by john santu last updated on 09/Feb/20
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Commented by MJS last updated on 09/Feb/20
...this leads to a polynome of 6^(th) degree: t^6 −t^5 −5t^4 −2t^3 +t^2 +3t−1=0 which can be factorized... (t^2 +t−1)(t^2 −(1+(√5))t+1)(t^2 −(1−(√5))t+1)=0 ...and solved
$$…\mathrm{this}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{a}\:\mathrm{polynome}\:\mathrm{of}\:\mathrm{6}^{\mathrm{th}} \:\mathrm{degree}: \\ $$$${t}^{\mathrm{6}} −{t}^{\mathrm{5}} −\mathrm{5}{t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{3}} +{t}^{\mathrm{2}} +\mathrm{3}{t}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{which}\:\mathrm{can}\:\mathrm{be}\:\mathrm{factorized}… \\ $$$$\left({t}^{\mathrm{2}} +{t}−\mathrm{1}\right)\left({t}^{\mathrm{2}} −\left(\mathrm{1}+\sqrt{\mathrm{5}}\right){t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} −\left(\mathrm{1}−\sqrt{\mathrm{5}}\right){t}+\mathrm{1}\right)=\mathrm{0} \\ $$$$…\mathrm{and}\:\mathrm{solved} \\ $$
Commented by john santu last updated on 09/Feb/20
my way sir (1)cos^2 x(1+sin x−2cos x)=cos^2 x−2sin xcos x+sin^2 x (2)1+sin x−2cos x≥0 (3)cos x(cos x−sin x)>0 ⇒(1) (sin x−2cos x)(cos^2 x−sin x)=0 (i) sin x=2cos x ⇒tan x=2 (ii) cos^2 x−sin x=0 sin^2 x+sin x−1=0 ⇒sin x=((−1+(√5))/2) x=sin^(−1) (((−1+(√5))/2))+2πn , n∈Z x=π−sin^(−1) (((−1+(√5))/2))+2πn, n∈Z (3)cos x(cos x−sin x)>0 cos x=0 ⇒x=(π/2)+πn, n∈Z cos x=sin x ⇒x=(π/4)+πk ,k∈Z
$${my}\:{way}\:{sir}\: \\ $$$$\left(\mathrm{1}\right)\mathrm{cos}\:^{\mathrm{2}} {x}\left(\mathrm{1}+\mathrm{sin}\:{x}−\mathrm{2cos}\:{x}\right)=\mathrm{cos}\:^{\mathrm{2}} {x}−\mathrm{2sin}\:{x}\mathrm{cos}\:{x}+\mathrm{sin}\:^{\mathrm{2}} {x} \\ $$$$\left(\mathrm{2}\right)\mathrm{1}+\mathrm{sin}\:{x}−\mathrm{2cos}\:{x}\geqslant\mathrm{0} \\ $$$$\left(\mathrm{3}\right)\mathrm{cos}\:{x}\left(\mathrm{cos}\:{x}−\mathrm{sin}\:{x}\right)>\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{1}\right)\:\left(\mathrm{sin}\:{x}−\mathrm{2cos}\:{x}\right)\left(\mathrm{cos}\:^{\mathrm{2}} {x}−\mathrm{sin}\:{x}\right)=\mathrm{0} \\ $$$$\left({i}\right)\:\mathrm{sin}\:{x}=\mathrm{2cos}\:{x}\:\Rightarrow\mathrm{tan}\:{x}=\mathrm{2} \\ $$$$\left({ii}\right)\:\mathrm{cos}\:^{\mathrm{2}} {x}−\mathrm{sin}\:{x}=\mathrm{0} \\ $$$$\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{sin}\:{x}−\mathrm{1}=\mathrm{0}\:\Rightarrow\mathrm{sin}\:{x}=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${x}=\mathrm{sin}^{−\mathrm{1}} \left(\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)+\mathrm{2}\pi{n}\:,\:{n}\in\mathbb{Z} \\ $$$${x}=\pi−\mathrm{sin}^{−\mathrm{1}} \left(\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)+\mathrm{2}\pi{n},\:{n}\in\mathbb{Z} \\ $$$$\left(\mathrm{3}\right)\mathrm{cos}\:{x}\left(\mathrm{cos}\:{x}−\mathrm{sin}\:{x}\right)>\mathrm{0} \\ $$$$\mathrm{cos}\:{x}=\mathrm{0}\:\Rightarrow{x}=\frac{\pi}{\mathrm{2}}+\pi{n},\:{n}\in\mathbb{Z} \\ $$$$\mathrm{cos}\:{x}=\mathrm{sin}\:{x}\:\Rightarrow{x}=\frac{\pi}{\mathrm{4}}+\pi{k}\:,{k}\in\mathbb{Z} \\ $$$$ \\ $$

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