Menu Close

Solve-sinx-cosx-1-x-R-

Tinku Tara 0 Comments
Pin




Question Number 14189 by tawa tawa last updated on 29/May/17
Solve: (√(sinx)) + (√(cosx)) = 1 x ∈ R
$$\mathrm{Solve}:\:\:\:\sqrt{\mathrm{sinx}}\:+\:\sqrt{\mathrm{cosx}}\:=\:\mathrm{1} \\ $$$$\mathrm{x}\:\in\:\mathrm{R} \\ $$
Answered by ajfour last updated on 29/May/17
let (√(sin x))=u and (√(cos x))=v this means u+v=1 ...(i) further as, sin^2 x+cos^2 x=1 u^4 +v^4 =1 ...(ii) we should know (u+v)^4 =u^4 +v^4 +uv(u^2 +v^2 ) +3uv(u+v)^2 using (i) and (ii): 1=1+uv(1−2uv)+3uv ⇒ uv(1−2uv+3)=0 uv=0 , or uv=2 since (√(sin x))(√(cos x))≠2, so uv=(√(sin x))(√(cos x))=0 ⇒ either sin x=0, or cos x=0 and since here sin x≥0 , cos x≥0 x=2nπ or x=2nπ+(π/2) .
$${let}\:\sqrt{\mathrm{sin}\:{x}}={u}\:\:{and}\:\sqrt{\mathrm{cos}\:{x}}={v} \\ $$$${this}\:{means}\:\:{u}+{v}=\mathrm{1}\:\:\:\:\:\:\:…\left({i}\right) \\ $$$${further}\:{as},\:\:\:\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{cos}\:^{\mathrm{2}} {x}=\mathrm{1} \\ $$$$\:\:\:\:{u}^{\mathrm{4}} +{v}^{\mathrm{4}} =\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\left({ii}\right) \\ $$$${we}\:{should}\:{know} \\ $$$$\:\:\:\left({u}+{v}\right)^{\mathrm{4}} ={u}^{\mathrm{4}} +{v}^{\mathrm{4}} +{uv}\left({u}^{\mathrm{2}} +{v}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{3}{uv}\left({u}+{v}\right)^{\mathrm{2}} \\ $$$${using}\:\left({i}\right)\:{and}\:\left({ii}\right): \\ $$$$\:\:\mathrm{1}=\mathrm{1}+{uv}\left(\mathrm{1}−\mathrm{2}{uv}\right)+\mathrm{3}{uv} \\ $$$$\Rightarrow\:\:{uv}\left(\mathrm{1}−\mathrm{2}{uv}+\mathrm{3}\right)=\mathrm{0} \\ $$$${uv}=\mathrm{0}\:\:,\:{or}\:\:{uv}=\mathrm{2} \\ $$$${since}\:\:\:\sqrt{\mathrm{sin}\:{x}}\sqrt{\mathrm{cos}\:{x}}\neq\mathrm{2},\:\:\:{so} \\ $$$${uv}=\sqrt{\mathrm{sin}\:{x}}\sqrt{\mathrm{cos}\:{x}}=\mathrm{0} \\ $$$$\Rightarrow\:{either}\:\:\mathrm{sin}\:{x}=\mathrm{0},\:{or}\:\mathrm{cos}\:{x}=\mathrm{0}\: \\ $$$${and}\:{since}\:{here}\:\mathrm{sin}\:{x}\geqslant\mathrm{0}\:,\:\mathrm{cos}\:{x}\geqslant\mathrm{0}\: \\ $$$$\:\:{x}=\mathrm{2}{n}\pi\: \\ $$$${or}\:\:{x}=\mathrm{2}{n}\pi+\frac{\pi}{\mathrm{2}}\:. \\ $$
Commented by tawa tawa last updated on 29/May/17
Wow, God bless you sir.
$$\mathrm{Wow},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by mrW1 last updated on 29/May/17
x=2kπ or 2kπ+(π/2)
$${x}=\mathrm{2}{k}\pi\:{or}\:\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{2}} \\ $$
Commented by tawa tawa last updated on 29/May/17
Sir. there appears to be a contradition from line 5 to 4 from the bottom.
$$\mathrm{Sir}.\:\mathrm{there}\:\mathrm{appears}\:\mathrm{to}\:\mathrm{be}\:\mathrm{a}\:\mathrm{contradition}\:\mathrm{from}\:\mathrm{line}\:\mathrm{5}\:\mathrm{to}\:\mathrm{4}\:\mathrm{from}\:\mathrm{the}\:\mathrm{bottom}. \\ $$
Commented by tawa tawa last updated on 29/May/17
Sir. can you explain ???.or it is correct !
$$\mathrm{Sir}.\:\mathrm{can}\:\mathrm{you}\:\mathrm{explain}\:???.\mathrm{or}\:\mathrm{it}\:\mathrm{is}\:\mathrm{correct}\:! \\ $$
Commented by mrW1 last updated on 29/May/17
if x=((nπ)/2) and n=3 cos x=−1 (√(cos x))=(√(−1)) !
$${if}\:{x}=\frac{{n}\pi}{\mathrm{2}}\:{and}\:{n}=\mathrm{3} \\ $$$$\mathrm{cos}\:{x}=−\mathrm{1} \\ $$$$\sqrt{\mathrm{cos}\:{x}}=\sqrt{−\mathrm{1}}\:\:! \\ $$
Commented by ajfour last updated on 29/May/17
very sorry .. i was elsewhere in my thoughts..
$${very}\:{sorry}\:..\:{i}\:{was}\:{elsewhere}\:{in} \\ $$$${my}\:{thoughts}.. \\ $$
Commented by tawa tawa last updated on 29/May/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by tawa tawa last updated on 29/May/17
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by ajfour last updated on 29/May/17
Miss Tawa Q. 14194 upload again please..
$${Miss}\:{Tawa}\:{Q}.\:\mathrm{14194}\:{upload} \\ $$$${again}\:{please}.. \\ $$
Commented by tawa tawa last updated on 29/May/17
Have done that sir.
$$\mathrm{Have}\:\mathrm{done}\:\mathrm{that}\:\mathrm{sir}. \\ $$
Commented by tawa tawa last updated on 29/May/17
Where have you been sir MRW1
$$\mathrm{Where}\:\mathrm{have}\:\mathrm{you}\:\mathrm{been}\:\mathrm{sir}\:\mathrm{MRW1} \\ $$
Commented by mrW1 last updated on 30/May/17
Dear Tawa: I′m still here. In recent time I watched more than give answers.
$${Dear}\:{Tawa}:\:{I}'{m}\:{still}\:{here}.\:{In}\:{recent} \\ $$$${time}\:{I}\:{watched}\:{more}\:{than}\:{give}\:{answers}. \\ $$
Commented by ajfour last updated on 09/Jun/18
Glad to know that, Sir.
$${Glad}\:{to}\:{know}\:{that},\:{Sir}. \\ $$
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 29/May/17
sinx+cosx+2(√(sinx.cosx))=1 2(√(sinx.cosx))=1−(sinx+cosx) 4sinx.cosx=1+(sin^2 x+cos^2 x+2sinx.cosx)−2(sinx+cosx) ⇒sinx+cosx=1−sinx.cosx 1+2sinx.cosx=1−2sinx.cosx+sin^2 x.cos^2 x ⇒sinx.cosx(sinx.cosx−4)=0 ⇒ { ((sinx.cosx=4⇒sin2x=8 (impossible))),((sinx.cosx=0⇒sin2x=0⇒2x=2kπ)) :} ⇒x=kπ .■
$${sinx}+{cosx}+\mathrm{2}\sqrt{{sinx}.{cosx}}=\mathrm{1} \\ $$$$\mathrm{2}\sqrt{{sinx}.{cosx}}=\mathrm{1}−\left({sinx}+{cosx}\right) \\ $$$$\mathrm{4}{sinx}.{cosx}=\mathrm{1}+\left({sin}^{\mathrm{2}} {x}+{cos}^{\mathrm{2}} {x}+\mathrm{2}{sinx}.{cosx}\right)−\mathrm{2}\left({sinx}+{cosx}\right) \\ $$$$\Rightarrow{sinx}+{cosx}=\mathrm{1}−{sinx}.{cosx} \\ $$$$\mathrm{1}+\mathrm{2}{sinx}.{cosx}=\mathrm{1}−\mathrm{2}{sinx}.{cosx}+{sin}^{\mathrm{2}} {x}.{cos}^{\mathrm{2}} {x} \\ $$$$\Rightarrow{sinx}.{cosx}\left({sinx}.{cosx}−\mathrm{4}\right)=\mathrm{0} \\ $$$$\Rightarrow\begin{cases}{{sinx}.{cosx}=\mathrm{4}\Rightarrow{sin}\mathrm{2}{x}=\mathrm{8}\:\:\left({impossible}\right)}\\{{sinx}.{cosx}=\mathrm{0}\Rightarrow{sin}\mathrm{2}{x}=\mathrm{0}\Rightarrow\mathrm{2}{x}=\mathrm{2}{k}\pi}\end{cases} \\ $$$$\Rightarrow{x}={k}\pi\:\:.\blacksquare \\ $$
Commented by tawa tawa last updated on 29/May/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by mrW1 last updated on 29/May/17
sin 2x=0 but cos x≥0 ⇒x=2kπ and 2kπ+(π/2)
$$\mathrm{sin}\:\mathrm{2}{x}=\mathrm{0} \\ $$$${but}\:\mathrm{cos}\:{x}\geqslant\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{2}{k}\pi\:{and}\:\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{2}} \\ $$
Commented by tawa tawa last updated on 29/May/17
Where did you go sir MrW1, have not been seeing you online.
$$\mathrm{Where}\:\mathrm{did}\:\mathrm{you}\:\mathrm{go}\:\mathrm{sir}\:\mathrm{MrW1},\:\mathrm{have}\:\mathrm{not}\:\mathrm{been}\:\mathrm{seeing}\:\mathrm{you}\:\mathrm{online}. \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *