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2sin-2x-y-sin-y-cos-2x-sin-2x-sin-2y-2-Find-the-solution-

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Question Number 197437 by dimentri last updated on 17/Sep/23
{ ((2sin (2x+y) sin y = cos 2x)),((sin 2x−sin 2y=(√2))) :} Find the solution
$$\:\:\:\:\begin{cases}{\mathrm{2sin}\:\left(\mathrm{2}{x}+{y}\right)\:\mathrm{sin}\:{y}\:=\:\mathrm{cos}\:\mathrm{2}{x}}\\{\mathrm{sin}\:\mathrm{2}{x}−\mathrm{sin}\:\mathrm{2}{y}=\sqrt{\mathrm{2}}}\end{cases} \\ $$$$\:\:{Find}\:{the}\:{solution} \\ $$
Answered by Frix last updated on 17/Sep/23
x=tan^(−1) u ∧y=tan^(−1) v The equations are then easy to solve ⇒ x=n_1 π+((3π)/8)∧y=n_2 π−(π/8) ∨ x=n_1 π+(π/8)∧y=n_2 π−((3π)/8) For 0≤x, y<π x=((3π)/8)∧y=((7π)/8)∨x=(π/8)∧y=((5π)/8)
$${x}=\mathrm{tan}^{−\mathrm{1}} \:{u}\:\wedge{y}=\mathrm{tan}^{−\mathrm{1}} \:{v} \\ $$$$\mathrm{The}\:\mathrm{equations}\:\mathrm{are}\:\mathrm{then}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve} \\ $$$$\Rightarrow \\ $$$${x}={n}_{\mathrm{1}} \pi+\frac{\mathrm{3}\pi}{\mathrm{8}}\wedge{y}={n}_{\mathrm{2}} \pi−\frac{\pi}{\mathrm{8}} \\ $$$$\vee \\ $$$${x}={n}_{\mathrm{1}} \pi+\frac{\pi}{\mathrm{8}}\wedge{y}={n}_{\mathrm{2}} \pi−\frac{\mathrm{3}\pi}{\mathrm{8}} \\ $$$$\mathrm{For}\:\mathrm{0}\leqslant{x},\:{y}<\pi \\ $$$${x}=\frac{\mathrm{3}\pi}{\mathrm{8}}\wedge{y}=\frac{\mathrm{7}\pi}{\mathrm{8}}\vee{x}=\frac{\pi}{\mathrm{8}}\wedge{y}=\frac{\mathrm{5}\pi}{\mathrm{8}} \\ $$
Answered by cortano12 last updated on 17/Sep/23
solution is x = { (π/8)+k.π ; ((3π)/8)+ℓ.π } y = {−(π/8)+k.π ; ((5π)/8)+ℓ.π }
$$\:\mathrm{solution}\:\mathrm{is}\: \\ $$$$\:\:\:\mathrm{x}\:=\:\left\{\:\frac{\pi}{\mathrm{8}}+\mathrm{k}.\pi\:;\:\frac{\mathrm{3}\pi}{\mathrm{8}}+\ell.\pi\:\right\} \\ $$$$\:\:\:\mathrm{y}\:=\:\left\{−\frac{\pi}{\mathrm{8}}+\mathrm{k}.\pi\:;\:\frac{\mathrm{5}\pi}{\mathrm{8}}+\ell.\pi\:\right\} \\ $$
Answered by MM42 last updated on 17/Sep/23
cos2x−cos2(x+y)=cos2x⇒2(x+y)=kπ+(π/2) 2cos^2 (x+y)−1=0⇒cos(x+y)=±((√2)/2) ⇒x+y=kπ+(π/4) ; ...,−((3π)/4),(π/4),((5π)/4),... 2sin(x−y)cos(x+y)=(√2) ⇒sin(x−y)=±1 ⇒x−y=kπ+(π/2) ; −(π/2) ,(π/2),((3π)/2),... ⇒x=kπ+((3π)/8) & y=kπ−(π/8)
$${cos}\mathrm{2}{x}−{cos}\mathrm{2}\left({x}+{y}\right)={cos}\mathrm{2}{x}\Rightarrow\mathrm{2}\left({x}+{y}\right)={k}\pi+\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{2}{cos}^{\mathrm{2}} \left({x}+{y}\right)−\mathrm{1}=\mathrm{0}\Rightarrow{cos}\left({x}+{y}\right)=\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\Rightarrow{x}+{y}={k}\pi+\frac{\pi}{\mathrm{4}}\:;\:…,−\frac{\mathrm{3}\pi}{\mathrm{4}},\frac{\pi}{\mathrm{4}},\frac{\mathrm{5}\pi}{\mathrm{4}},… \\ $$$$\mathrm{2}{sin}\left({x}−{y}\right){cos}\left({x}+{y}\right)=\sqrt{\mathrm{2}} \\ $$$$\Rightarrow{sin}\left({x}−{y}\right)=\pm\mathrm{1} \\ $$$$\Rightarrow{x}−{y}={k}\pi+\frac{\pi}{\mathrm{2}}\:;\:−\frac{\pi}{\mathrm{2}}\:,\frac{\pi}{\mathrm{2}},\frac{\mathrm{3}\pi}{\mathrm{2}},… \\ $$$$\:\Rightarrow{x}={k}\pi+\frac{\mathrm{3}\pi}{\mathrm{8}}\:\:\:\:\:\:\&\:\:{y}={k}\pi−\frac{\pi}{\mathrm{8}}\:\: \\ $$$$ \\ $$
Answered by witcher3 last updated on 17/Sep/23
2sin(2x+y)sin(y) cos(a)−cos(b)=−2sin(((a−b)/2))sin(((a+b)/2)) ⇒2sn(2x+y)sin(y)=cos(2x)−cos(2x+2y)=cos(2x) ⇒cos(2x+2y)=0⇒2x+2y=(π/2)+nπ sin(2x)−sin(2y)=(√2) ⇒sin((π/2)+nπ−2y)−sin(2y)=(√2) ⇒cos(nπ−2y)−sin(2y)=(√2) ⇒(−1)^n cos(2y)−sin(2y)=(√2) n=2k⇒ cos(2y)−sin(2y)=(√2) ⇒cos(2y+(π/4))=1 ⇒2y+(π/4)=2kπ⇒y=−(π/8)+mπ,x=((3π)/8)+kπ n=2k+1 2y−(π/4)=π+2kπ ⇒y=((5π)/8)+kπ x=−(π/8)+nπ
$$\mathrm{2sin}\left(\mathrm{2x}+\mathrm{y}\right)\mathrm{sin}\left(\mathrm{y}\right) \\ $$$$\mathrm{cos}\left(\mathrm{a}\right)−\mathrm{cos}\left(\mathrm{b}\right)=−\mathrm{2sin}\left(\frac{\mathrm{a}−\mathrm{b}}{\mathrm{2}}\right)\mathrm{sin}\left(\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\mathrm{2sn}\left(\mathrm{2x}+\mathrm{y}\right)\mathrm{sin}\left(\mathrm{y}\right)=\mathrm{cos}\left(\mathrm{2x}\right)−\mathrm{cos}\left(\mathrm{2x}+\mathrm{2y}\right)=\mathrm{cos}\left(\mathrm{2x}\right) \\ $$$$\Rightarrow\mathrm{cos}\left(\mathrm{2}{x}+\mathrm{2}{y}\right)=\mathrm{0}\Rightarrow\mathrm{2x}+\mathrm{2y}=\frac{\pi}{\mathrm{2}}+\mathrm{n}\pi \\ $$$$\mathrm{sin}\left(\mathrm{2x}\right)−\mathrm{sin}\left(\mathrm{2y}\right)=\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{sin}\left(\frac{\pi}{\mathrm{2}}+\mathrm{n}\pi−\mathrm{2y}\right)−\mathrm{sin}\left(\mathrm{2y}\right)=\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{cos}\left(\mathrm{n}\pi−\mathrm{2y}\right)−\mathrm{sin}\left(\mathrm{2y}\right)=\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{cos}\left(\mathrm{2y}\right)−\mathrm{sin}\left(\mathrm{2y}\right)=\sqrt{\mathrm{2}} \\ $$$$\mathrm{n}=\mathrm{2k}\Rightarrow \\ $$$$\mathrm{cos}\left(\mathrm{2y}\right)−\mathrm{sin}\left(\mathrm{2y}\right)=\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{cos}\left(\mathrm{2y}+\frac{\pi}{\mathrm{4}}\right)=\mathrm{1} \\ $$$$\Rightarrow\mathrm{2y}+\frac{\pi}{\mathrm{4}}=\mathrm{2k}\pi\Rightarrow\mathrm{y}=−\frac{\pi}{\mathrm{8}}+\mathrm{m}\pi,\mathrm{x}=\frac{\mathrm{3}\pi}{\mathrm{8}}+\mathrm{k}\pi \\ $$$$\mathrm{n}=\mathrm{2k}+\mathrm{1} \\ $$$$\mathrm{2y}−\frac{\pi}{\mathrm{4}}=\pi+\mathrm{2k}\pi \\ $$$$\Rightarrow\mathrm{y}=\frac{\mathrm{5}\pi}{\mathrm{8}}+\mathrm{k}\pi \\ $$$$\mathrm{x}=−\frac{\pi}{\mathrm{8}}+\mathrm{n}\pi \\ $$

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