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Question Number 59441 by rahul 19 last updated on 10/May/19

Answered by tanmay last updated on 10/May/19

sinx+2costcosx=2 4cos^2 t(1−sin^2 x)=4−4sinx+sin^2 x cost=c sinx=k 4c^2 (1−k^2 )=4−4k+k^2 k^2 +4c^2 k^2 −4k+4−4c^2 =0 k^2 (1+4c^2 )+k(−4)+4−4c^2 =0 k=((4±(√(16−4×(1+4c^2 )(4−4c^2 ))))/(2(1+4c^2 ))) =((4±(√(16−4(4−4c^2 +16c^2 −16c^4 )))/(2(1+4c^2 ))) =((4±(√(16−16+16c^2 −64c^2 +64c^4 )))/(2(1+4c^2 ))) =((4±(√(64c^4 −48c^2 )))/(2(1+4c^2 ))) 64c^4 −48c^2 ≥0 16c^2 (4c^2 −3)≥0 16c^2 (2c+(√3) )(2c−(√3) )≥0 critcal value of c=0,((√3)/2),((−(√3))/2) f(c)=16c^2 (2c+(√3) )(2c−(√3) ) f(c)>0 when c>((√3)/2) f(c)>0 when c<((−(√3))/2) f(c)<0 when ((√3)/2)>c>0 f(c) <0 when 0> c>((−(√3))/2) cost=((√3)/2) →sint=(1/2) rahul pls check mistake if any

$${sinx}+\mathrm{2}{costcosx}=\mathrm{2} \\ $$$$\mathrm{4}{cos}^{\mathrm{2}} {t}\left(\mathrm{1}−{sin}^{\mathrm{2}} {x}\right)=\mathrm{4}−\mathrm{4}{sinx}+{sin}^{\mathrm{2}} {x} \\ $$$${cost}={c}\:\:\:{sinx}={k} \\ $$$$\mathrm{4}{c}^{\mathrm{2}} \left(\mathrm{1}−{k}^{\mathrm{2}} \right)=\mathrm{4}−\mathrm{4}{k}+{k}^{\mathrm{2}} \\ $$$${k}^{\mathrm{2}} +\mathrm{4}{c}^{\mathrm{2}} {k}^{\mathrm{2}} −\mathrm{4}{k}+\mathrm{4}−\mathrm{4}{c}^{\mathrm{2}} =\mathrm{0} \\ $$$${k}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{4}{c}^{\mathrm{2}} \right)+{k}\left(−\mathrm{4}\right)+\mathrm{4}−\mathrm{4}{c}^{\mathrm{2}} =\mathrm{0} \\ $$$${k}=\frac{\mathrm{4}\pm\sqrt{\mathrm{16}−\mathrm{4}×\left(\mathrm{1}+\mathrm{4}{c}^{\mathrm{2}} \right)\left(\mathrm{4}−\mathrm{4}{c}^{\mathrm{2}} \right)}}{\mathrm{2}\left(\mathrm{1}+\mathrm{4}{c}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{4}\pm\sqrt{\mathrm{16}−\mathrm{4}\left(\mathrm{4}−\mathrm{4}{c}^{\mathrm{2}} +\mathrm{16}{c}^{\mathrm{2}} −\mathrm{16}{c}^{\mathrm{4}} \right.}}{\mathrm{2}\left(\mathrm{1}+\mathrm{4}{c}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{4}\pm\sqrt{\mathrm{16}−\mathrm{16}+\mathrm{16}{c}^{\mathrm{2}} −\mathrm{64}{c}^{\mathrm{2}} +\mathrm{64}{c}^{\mathrm{4}} }}{\mathrm{2}\left(\mathrm{1}+\mathrm{4}{c}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{4}\pm\sqrt{\mathrm{64}{c}^{\mathrm{4}} −\mathrm{48}{c}^{\mathrm{2}} }}{\mathrm{2}\left(\mathrm{1}+\mathrm{4}{c}^{\mathrm{2}} \right)} \\ $$$$\mathrm{64}{c}^{\mathrm{4}} −\mathrm{48}{c}^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\mathrm{16}{c}^{\mathrm{2}} \left(\mathrm{4}{c}^{\mathrm{2}} −\mathrm{3}\right)\geqslant\mathrm{0} \\ $$$$\mathrm{16}{c}^{\mathrm{2}} \left(\mathrm{2}{c}+\sqrt{\mathrm{3}}\:\right)\left(\mathrm{2}{c}−\sqrt{\mathrm{3}}\:\right)\geqslant\mathrm{0} \\ $$$${critcal}\:{value}\:{of}\:{c}=\mathrm{0},\frac{\sqrt{\mathrm{3}}}{\mathrm{2}},\frac{−\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${f}\left({c}\right)=\mathrm{16}{c}^{\mathrm{2}} \left(\mathrm{2}{c}+\sqrt{\mathrm{3}}\:\right)\left(\mathrm{2}{c}−\sqrt{\mathrm{3}}\:\right) \\ $$$${f}\left({c}\right)>\mathrm{0}\:\:{when}\:{c}>\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${f}\left({c}\right)>\mathrm{0}\:\:{when}\:{c}<\frac{−\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${f}\left({c}\right)<\mathrm{0}\:\:{when}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}>{c}>\mathrm{0} \\ $$$${f}\left({c}\right)\:<\mathrm{0}\:\:\:\:\:\:{when}\:\mathrm{0}>\:{c}>\frac{−\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${cost}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:\rightarrow{sint}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${rahul}\:{pls}\:{check}\:{mistake}\:{if}\:{any} \\ $$

Commented by rahul 19 last updated on 10/May/19

thank u sir.

$${thank}\:{u}\:{sir}. \\ $$

Commented by tanmay last updated on 10/May/19

i want to share some informationfor you...pls reply if you see this comment...

$${i}\:{want}\:{to}\:{share}\:{some}\:{informationfor}\:{you}...{pls} \\ $$$${reply}\:{if}\:{you}\:{see}\:{this}\:{comment}... \\ $$

Commented by rahul 19 last updated on 10/May/19

Sure,sir.

$${Sure},{sir}. \\ $$

Commented by tanmay last updated on 10/May/19

i have found another platform where you can ask your doubt about phy+chem+math and doubt is cleared by others just like this platform but in that platform mostly students studying for IIT indian students... if you want i can share...

$${i}\:{have}\:{found}\:{another}\:{platform}\:{where}\:{you}\:{can} \\ $$$${ask}\:{your}\:{doubt}\:{about}\:{phy}+{chem}+{math}\:{and} \\ $$$${doubt}\:{is}\:{cleared}\:{by}\:{others}\:{just}\:{like}\:{this}\:{platform} \\ $$$${but}\:{in}\:{that}\:{platform}\:{mostly}\:{students}\:{studying} \\ $$$${for}\:{IIT}\:{indian}\:{students}... \\ $$$${if}\:{you}\:{want}\:{i}\:{can}\:{share}... \\ $$

Commented by rahul 19 last updated on 10/May/19

i′m sorry, but it′s too late for me now. just 10−15 days more... Anyways , thanks for the concern !

$${i}'{m}\:{sorry},\:{but}\:{it}'{s}\:{too}\:{late}\:{for}\:{me}\:{now}. \\ $$$${just}\:\mathrm{10}−\mathrm{15}\:{days}\:{more}... \\ $$$${Anyways}\:,\:{thanks}\:{for}\:{the}\:{concern}\:! \\ $$

Commented by tanmay last updated on 13/May/19

website goiit.com

$${website}\:{goiit}.{com} \\ $$

Commented by necx1 last updated on 13/May/19

please share

$${please}\:{share} \\ $$

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