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2sin-2x-4sin-2-x-7cos-2x-with-pi-2-lt-x-lt-pi-find-sin-2x-

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Question Number 104576 by bemath last updated on 22/Jul/20
2sin 2x −4sin^2 x = 7cos 2x with (π/2)< x < π find sin 2x
$$\mathrm{2sin}\:\mathrm{2}{x}\:−\mathrm{4sin}\:^{\mathrm{2}} {x}\:=\:\mathrm{7cos}\:\mathrm{2}{x} \\ $$$${with}\:\frac{\pi}{\mathrm{2}}<\:{x}\:<\:\pi\: \\ $$$${find}\:\mathrm{sin}\:\mathrm{2}{x}\: \\ $$
Commented by bemath last updated on 22/Jul/20
thank you both
$${thank}\:{you}\:{both} \\ $$
Answered by bobhans last updated on 22/Jul/20
⇔4sin x cos x−4sin^2 x = 7(cos^2 x−sin^2 x) 4sin x (cos x−sin x) = 7(cos x−sin x)(cos x+sin x) ⇒(sin x−cos x){4sin x−7cos x−7sin x)=0 { ((sin x=cos x (rejected))),((7cos x = −3sin x )) :} tan x = −(7/3) → { ((sin x = (7/( (√(58)))))),((cos x = −(3/( (√(58)))))) :} sin 2x = 2×(7/( (√(58)))) ×−(3/( (√(58)))) = −((21)/(29)) ★
$$\Leftrightarrow\mathrm{4sin}\:{x}\:\mathrm{cos}\:{x}−\mathrm{4sin}\:^{\mathrm{2}} {x}\:=\:\mathrm{7}\left(\mathrm{cos}\:^{\mathrm{2}} {x}−\mathrm{sin}\:^{\mathrm{2}} {x}\right) \\ $$$$\mathrm{4sin}\:{x}\:\left(\mathrm{cos}\:{x}−\mathrm{sin}\:{x}\right)\:=\:\mathrm{7}\left(\mathrm{cos}\:{x}−\mathrm{sin}\:{x}\right)\left(\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\right) \\ $$$$\Rightarrow\left(\mathrm{sin}\:{x}−\mathrm{cos}\:{x}\right)\left\{\mathrm{4sin}\:{x}−\mathrm{7cos}\:{x}−\mathrm{7sin}\:{x}\right)=\mathrm{0} \\ $$$$\begin{cases}{\mathrm{sin}\:{x}=\mathrm{cos}\:{x}\:\left({rejected}\right)}\\{\mathrm{7cos}\:{x}\:=\:−\mathrm{3sin}\:{x}\:}\end{cases} \\ $$$$\mathrm{tan}\:{x}\:=\:−\frac{\mathrm{7}}{\mathrm{3}}\:\rightarrow\begin{cases}{\mathrm{sin}\:{x}\:=\:\frac{\mathrm{7}}{\:\sqrt{\mathrm{58}}}}\\{\mathrm{cos}\:{x}\:=\:−\frac{\mathrm{3}}{\:\sqrt{\mathrm{58}}}}\end{cases} \\ $$$$\mathrm{sin}\:\mathrm{2}{x}\:=\:\mathrm{2}×\frac{\mathrm{7}}{\:\sqrt{\mathrm{58}}}\:×−\frac{\mathrm{3}}{\:\sqrt{\mathrm{58}}}\:=\:−\frac{\mathrm{21}}{\mathrm{29}}\:\bigstar \\ $$
Answered by OlafThorendsen last updated on 22/Jul/20
2sin2x−4((1−cos2x)/2) = 7cos2x 2sin2x−2 = 5cos2x ⇒ 4sin^2 2x−8sin2x+4 = 25cos^2 2x 4sin^2 2x−8sin2x+4 = 25−25sin^2 2x 29sin^2 x−8sin2x−21 = 0 (sin2x−1)(29sin2x+21) = 0 sin2x = 1 ⇔ x = (π/4)+kπ, k∈Z impossible, (π/2)<x<π sin2x = −((21)/(29)) is a solution.
$$\mathrm{2sin2}{x}−\mathrm{4}\frac{\mathrm{1}−\mathrm{cos2}{x}}{\mathrm{2}}\:=\:\mathrm{7cos2}{x} \\ $$$$\mathrm{2sin2}{x}−\mathrm{2}\:=\:\mathrm{5cos2}{x} \\ $$$$\Rightarrow\:\mathrm{4sin}^{\mathrm{2}} \mathrm{2}{x}−\mathrm{8sin2}{x}+\mathrm{4}\:=\:\mathrm{25cos}^{\mathrm{2}} \mathrm{2}{x} \\ $$$$\mathrm{4sin}^{\mathrm{2}} \mathrm{2}{x}−\mathrm{8sin2}{x}+\mathrm{4}\:=\:\mathrm{25}−\mathrm{25sin}^{\mathrm{2}} \mathrm{2}{x} \\ $$$$\mathrm{29sin}^{\mathrm{2}} {x}−\mathrm{8sin2}{x}−\mathrm{21}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{sin2}{x}−\mathrm{1}\right)\left(\mathrm{29sin2}{x}+\mathrm{21}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{sin2}{x}\:=\:\mathrm{1}\:\Leftrightarrow\:{x}\:=\:\frac{\pi}{\mathrm{4}}+{k}\pi,\:{k}\in\mathbb{Z} \\ $$$$\mathrm{impossible},\:\frac{\pi}{\mathrm{2}}<{x}<\pi \\ $$$$\mathrm{sin2}{x}\:=\:−\frac{\mathrm{21}}{\mathrm{29}}\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution}. \\ $$$$ \\ $$
Answered by Dwaipayan Shikari last updated on 22/Jul/20
2sin2x−4sin^2 x=7cos^2 x−7sin^2 x −7cos^2 x+4sinxcosx+3sin^2 x=0 −7cos^2 x+7sinxcosx−3sinxcosx+3sin^2 x=0 −7cosx(cosx−sinx)−3sinx(cosx−sinx)=0 (cosx−sinx)(7cosx+3sinx)=0 cosx=sinx tanx=1 ((2tanx)/(tan^2 x+1))=sin2x=1(But it is not the solution) 7cosx=−3sinx tanx=((−7)/3) ((2tanx)/(tan^2 x+1))=(((−14)/3)/(((49)/9)+1))=((−14.3)/(58))=−((21)/(29))
$$\mathrm{2}{sin}\mathrm{2}{x}−\mathrm{4}{sin}^{\mathrm{2}} {x}=\mathrm{7}{cos}^{\mathrm{2}} {x}−\mathrm{7}{sin}^{\mathrm{2}} {x} \\ $$$$−\mathrm{7}{cos}^{\mathrm{2}} {x}+\mathrm{4}{sinxcosx}+\mathrm{3}{sin}^{\mathrm{2}} {x}=\mathrm{0} \\ $$$$−\mathrm{7}{cos}^{\mathrm{2}} {x}+\mathrm{7}{sinxcosx}−\mathrm{3}{sinxcosx}+\mathrm{3}{sin}^{\mathrm{2}} {x}=\mathrm{0} \\ $$$$−\mathrm{7}{cosx}\left({cosx}−{sinx}\right)−\mathrm{3}{sinx}\left({cosx}−{sinx}\right)=\mathrm{0} \\ $$$$\left({cosx}−{sinx}\right)\left(\mathrm{7}{cosx}+\mathrm{3}{sinx}\right)=\mathrm{0} \\ $$$${cosx}={sinx} \\ $$$${tanx}=\mathrm{1} \\ $$$$\frac{\mathrm{2}{tanx}}{{tan}^{\mathrm{2}} {x}+\mathrm{1}}={sin}\mathrm{2}{x}=\mathrm{1}\left({But}\:{it}\:{is}\:{not}\:{the}\:{solution}\right) \\ $$$$\mathrm{7}{cosx}=−\mathrm{3}{sinx} \\ $$$${tanx}=\frac{−\mathrm{7}}{\mathrm{3}} \\ $$$$\frac{\mathrm{2}{tanx}}{{tan}^{\mathrm{2}} {x}+\mathrm{1}}=\frac{\frac{−\mathrm{14}}{\mathrm{3}}}{\frac{\mathrm{49}}{\mathrm{9}}+\mathrm{1}}=\frac{−\mathrm{14}.\mathrm{3}}{\mathrm{58}}=−\frac{\mathrm{21}}{\mathrm{29}} \\ $$
Answered by ajfour last updated on 22/Jul/20
let cos 2x=t ⇒ 4(1−t^2 )=(7t+2−2t)^2 4−4t^2 =25t^2 +20t+4 ⇒ 29t^2 +20t=0 ⇒ t=−((20)/(29)) , sin 2x = −(√(1−t^2 )) sin 2x = −((√(49×9))/(29)) = −((21)/(29)) .
$${let}\:\:\mathrm{cos}\:\mathrm{2}{x}={t} \\ $$$$\Rightarrow\:\mathrm{4}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)=\left(\mathrm{7}{t}+\mathrm{2}−\mathrm{2}{t}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}−\mathrm{4}{t}^{\mathrm{2}} =\mathrm{25}{t}^{\mathrm{2}} +\mathrm{20}{t}+\mathrm{4} \\ $$$$\Rightarrow\:\:\mathrm{29}{t}^{\mathrm{2}} +\mathrm{20}{t}=\mathrm{0} \\ $$$$\Rightarrow\:\:{t}=−\frac{\mathrm{20}}{\mathrm{29}}\:,\:\:\mathrm{sin}\:\mathrm{2}{x}\:=\:−\sqrt{\mathrm{1}−{t}^{\mathrm{2}} } \\ $$$$\:\:\:\mathrm{sin}\:\mathrm{2}{x}\:=\:−\frac{\sqrt{\mathrm{49}×\mathrm{9}}}{\mathrm{29}}\:=\:−\frac{\mathrm{21}}{\mathrm{29}}\:. \\ $$

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