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Find-the-real-roots-of-the-equation-cos-4-x-sin-7-x-1-in-the-interval-pi-pi-

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Question Number 14977 by Tinkutara last updated on 06/Jun/17
Find the real roots of the equation cos^4 x + sin^7 x = 1 in the interval [−π, π].
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{real}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{cos}^{\mathrm{4}} \:{x}\:+\:\mathrm{sin}^{\mathrm{7}} \:{x}\:=\:\mathrm{1}\:\mathrm{in}\:\mathrm{the}\:\mathrm{interval}\:\left[−\pi,\:\pi\right]. \\ $$
Commented by mrW1 last updated on 06/Jun/17
cos x=±1 and sin x=0 or cos x=0 and sin x=1 ⇒x=−π,0,(π/2),π
$$\mathrm{cos}\:{x}=\pm\mathrm{1}\:{and}\:\mathrm{sin}\:{x}=\mathrm{0} \\ $$$${or} \\ $$$$\mathrm{cos}\:{x}=\mathrm{0}\:{and}\:\mathrm{sin}\:{x}=\mathrm{1} \\ $$$$\Rightarrow{x}=−\pi,\mathrm{0},\frac{\pi}{\mathrm{2}},\pi \\ $$
Commented by Tinkutara last updated on 07/Jun/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$
Answered by myintkhaing last updated on 07/Jun/17
sin^7 x = 1 −cos^4 x sin^7 x = sin^2 x (2 − sin^2 x) sin^2 x(sin^5 x+sin^2 x −2) = 0 sin^2 x = 0 or sin^5 x+sin^2 x−2=0 sin x = 0 ⇒ x = 0, ±π sin^5 x+sin^2 x−2=0 , here −1≤sin x≤1 if sin x = −1, L.H.S ≠ R.H.S if sin x = 1, L.H.S = R.H.S x = (π/2) x = 0,±π,(π/2)
$${sin}^{\mathrm{7}} \:{x}\:=\:\mathrm{1}\:−{cos}^{\mathrm{4}} \:{x} \\ $$$${sin}^{\mathrm{7}} \:{x}\:=\:{sin}^{\mathrm{2}} \:{x}\:\left(\mathrm{2}\:−\:{sin}^{\mathrm{2}} {x}\right) \\ $$$${sin}^{\mathrm{2}} \:{x}\left({sin}^{\mathrm{5}} \:{x}+{sin}^{\mathrm{2}} \:{x}\:−\mathrm{2}\right)\:=\:\mathrm{0} \\ $$$${sin}^{\mathrm{2}} \:{x}\:=\:\mathrm{0}\:{or}\:{sin}^{\mathrm{5}} {x}+{sin}^{\mathrm{2}} {x}−\mathrm{2}=\mathrm{0} \\ $$$${sin}\:{x}\:=\:\mathrm{0}\:\Rightarrow\:{x}\:=\:\mathrm{0},\:\pm\pi \\ $$$${sin}^{\mathrm{5}} {x}+{sin}^{\mathrm{2}} {x}−\mathrm{2}=\mathrm{0}\:,\:{here}\:−\mathrm{1}\leqslant{sin}\:{x}\leqslant\mathrm{1} \\ $$$${if}\:{sin}\:{x}\:=\:−\mathrm{1},\:{L}.{H}.{S}\:\neq\:{R}.{H}.{S} \\ $$$${if}\:{sin}\:{x}\:=\:\mathrm{1},\:{L}.{H}.{S}\:=\:{R}.{H}.{S} \\ $$$${x}\:=\:\frac{\pi}{\mathrm{2}} \\ $$$${x}\:=\:\mathrm{0},\pm\pi,\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$
Commented by Tinkutara last updated on 07/Jun/17
Thanks Sir! But why you reject the 2^(nd) factor?
$$\mathrm{Thanks}\:\mathrm{Sir}!\:\mathrm{But}\:\mathrm{why}\:\mathrm{you}\:\mathrm{reject}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \\ $$$$\mathrm{factor}? \\ $$
Commented by myintkhaing last updated on 07/Jun/17
My post is edited.
$${My}\:{post}\:{is}\:{edited}. \\ $$
Commented by Tinkutara last updated on 07/Jun/17
But how to identify this that it has no real root?
$$\mathrm{But}\:\mathrm{how}\:\mathrm{to}\:\mathrm{identify}\:\mathrm{this}\:\mathrm{that}\:\mathrm{it}\:\mathrm{has}\:\mathrm{no} \\ $$$$\mathrm{real}\:\mathrm{root}? \\ $$
Commented by Tinkutara last updated on 07/Jun/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$
Answered by mrW1 last updated on 07/Jun/17
This question can be generalised to cos^(2m) x + sin^(2n+1) x = 1 with m,n∈N^+ the answer is the same, independent from m and n. sin^(2n+1) x=1−cos^(2m) x≥0 and ≤1 ⇒0≤sin x≤1 for sin x=0 we get cos^(2m) x=1 or cos x=±1, this is possible. for sin x=1 we get cos^(2m) x=0 or cos x=0, this is possible. for 0<sin x<1 we get cos^(2m) x+sin^(2n+1) x=1 (cos^2 x)^m +sin^(2n+1) x=1 (1−sin^2 x)^m +sin^(2n+1) x=1 let t=sin x, 0<t<1 ⇒(1−t^2 )^m +t^(2n+1) =1 (1−t^2 )^m =1−t^(2n+1) =(1−t)(1+t+t^2 +...+t^(2n) ) ⇒(((1−t^2 )^m )/(1−t))=1+t+t^2 +...+t^(2n) >1+t ⇒(((1−t^2 )^m )/((1−t)(1+t)))>1 ⇒(1−t^2 )^(m−1) >1 but 0<1−t^2 <1 ⇒ (1−t^2 )^(m−1) <1 this is a contradiction, which results from the assumption that 0<sin x<1. i.e. the assumption is wrong. therefore the solution of the equation cos^(2m) x + sin^(2n+1) x = 1 with m,n∈N^+ is only sin x=0 and cos x=±1 ⇒x=kπ, k∈Z or sin x=1 and cos x=0 ⇒x=kπ+(−1)^k (π/2), k∈Z
$$\mathrm{This}\:\mathrm{question}\:\mathrm{can}\:\mathrm{be}\:\mathrm{generalised}\:\mathrm{to} \\ $$$$\mathrm{cos}^{\mathrm{2m}} \:\mathrm{x}\:+\:\mathrm{sin}^{\mathrm{2n}+\mathrm{1}} \:\mathrm{x}\:=\:\mathrm{1}\:\mathrm{with}\:\mathrm{m},\mathrm{n}\in\mathbb{N}^{+} \\ $$$$\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{the}\:\mathrm{same},\:\mathrm{independent} \\ $$$$\mathrm{from}\:\mathrm{m}\:\mathrm{and}\:\mathrm{n}. \\ $$$$\mathrm{sin}^{\mathrm{2n}+\mathrm{1}} \:\mathrm{x}=\mathrm{1}−\mathrm{cos}^{\mathrm{2m}} \:\mathrm{x}\geqslant\mathrm{0}\:\mathrm{and}\:\leqslant\mathrm{1} \\ $$$$\Rightarrow\mathrm{0}\leqslant\mathrm{sin}\:\mathrm{x}\leqslant\mathrm{1} \\ $$$$ \\ $$$$\mathrm{for}\:\mathrm{sin}\:\mathrm{x}=\mathrm{0}\:\mathrm{we}\:\mathrm{get}\:\mathrm{cos}^{\mathrm{2m}} \:\mathrm{x}=\mathrm{1}\:\mathrm{or} \\ $$$$\mathrm{cos}\:\mathrm{x}=\pm\mathrm{1},\:\mathrm{this}\:\mathrm{is}\:\mathrm{possible}. \\ $$$$ \\ $$$$\mathrm{for}\:\mathrm{sin}\:\mathrm{x}=\mathrm{1}\:\mathrm{we}\:\mathrm{get}\:\mathrm{cos}^{\mathrm{2m}} \:\mathrm{x}=\mathrm{0}\:\mathrm{or} \\ $$$$\mathrm{cos}\:\mathrm{x}=\mathrm{0},\:\mathrm{this}\:\mathrm{is}\:\mathrm{possible}. \\ $$$$ \\ $$$$\mathrm{for}\:\mathrm{0}<\mathrm{sin}\:\mathrm{x}<\mathrm{1}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{cos}^{\mathrm{2m}} \:\mathrm{x}+\mathrm{sin}^{\mathrm{2n}+\mathrm{1}} \:\mathrm{x}=\mathrm{1} \\ $$$$\left(\mathrm{cos}^{\mathrm{2}} \:\mathrm{x}\right)^{\mathrm{m}} +\mathrm{sin}^{\mathrm{2n}+\mathrm{1}} \:\mathrm{x}=\mathrm{1} \\ $$$$\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\mathrm{x}\right)^{\mathrm{m}} +\mathrm{sin}^{\mathrm{2n}+\mathrm{1}} \:\mathrm{x}=\mathrm{1} \\ $$$$\mathrm{let}\:\mathrm{t}=\mathrm{sin}\:\mathrm{x},\:\mathrm{0}<\mathrm{t}<\mathrm{1} \\ $$$$\Rightarrow\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{m}} +\mathrm{t}^{\mathrm{2n}+\mathrm{1}} =\mathrm{1} \\ $$$$\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{m}} =\mathrm{1}−\mathrm{t}^{\mathrm{2n}+\mathrm{1}} =\left(\mathrm{1}−\mathrm{t}\right)\left(\mathrm{1}+\mathrm{t}+\mathrm{t}^{\mathrm{2}} +…+\mathrm{t}^{\mathrm{2n}} \right) \\ $$$$\Rightarrow\frac{\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{m}} }{\mathrm{1}−\mathrm{t}}=\mathrm{1}+\mathrm{t}+\mathrm{t}^{\mathrm{2}} +…+\mathrm{t}^{\mathrm{2n}} >\mathrm{1}+\mathrm{t} \\ $$$$\Rightarrow\frac{\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{m}} }{\left(\mathrm{1}−\mathrm{t}\right)\left(\mathrm{1}+\mathrm{t}\right)}>\mathrm{1} \\ $$$$\Rightarrow\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{m}−\mathrm{1}} >\mathrm{1} \\ $$$$\mathrm{but}\:\mathrm{0}<\mathrm{1}−\mathrm{t}^{\mathrm{2}} <\mathrm{1}\:\Rightarrow\:\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{m}−\mathrm{1}} <\mathrm{1} \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\mathrm{contradiction},\:\mathrm{which}\:\mathrm{results} \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{assumption}\:\mathrm{that}\:\mathrm{0}<\mathrm{sin}\:\mathrm{x}<\mathrm{1}. \\ $$$$\mathrm{i}.\mathrm{e}.\:\mathrm{the}\:\mathrm{assumption}\:\mathrm{is}\:\mathrm{wrong}. \\ $$$$ \\ $$$$\mathrm{therefore}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{cos}^{\mathrm{2m}} \:\mathrm{x}\:+\:\mathrm{sin}^{\mathrm{2n}+\mathrm{1}} \:\mathrm{x}\:=\:\mathrm{1}\:\mathrm{with}\:\mathrm{m},\mathrm{n}\in\mathbb{N}^{+} \\ $$$$\mathrm{is}\:\mathrm{only} \\ $$$$\mathrm{sin}\:\mathrm{x}=\mathrm{0}\:\mathrm{and}\:\mathrm{cos}\:\mathrm{x}=\pm\mathrm{1}\: \\ $$$$\Rightarrow\mathrm{x}=\mathrm{k}\pi,\:\mathrm{k}\in\mathbb{Z} \\ $$$$\mathrm{or} \\ $$$$\mathrm{sin}\:\mathrm{x}=\mathrm{1}\:\mathrm{and}\:\mathrm{cos}\:\mathrm{x}=\mathrm{0}\: \\ $$$$\Rightarrow\mathrm{x}=\mathrm{k}\pi+\left(−\mathrm{1}\right)^{\mathrm{k}} \frac{\pi}{\mathrm{2}},\:\mathrm{k}\in\mathbb{Z} \\ $$
Commented by Tinkutara last updated on 07/Jun/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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