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Question Number 197437 by dimentri last updated on 17/Sep/23

$$\{\begin{array}{l}2\sin (2x+y)\sin y=\cos 2x\\ \sin 2x-\sin 2y=\sqrt{2}\end{array}$$$$Findthesolution$$

Answered by Frix last updated on 17/Sep/23

$$x={\tan }^{-1}u\wedge y={\tan }^{-1}v$$$$\mathrm{The}\mathrm{equations}\mathrm{are}\mathrm{then}\mathrm{easy}\mathrm{to}\mathrm{solve}$$$$\Rightarrow$$$$x=n_1\pi +\frac{3\pi }{8}\wedge y=n_2\pi -\frac{\pi }{8}$$$$\vee$$$$x=n_1\pi +\frac{\pi }{8}\wedge y=n_2\pi -\frac{3\pi }{8}$$$$\mathrm{For}0⩽x,y<\pi$$$$x=\frac{3\pi }{8}\wedge y=\frac{7\pi }{8}\vee x=\frac{\pi }{8}\wedge y=\frac{5\pi }{8}$$

Answered by cortano12 last updated on 17/Sep/23

$$\mathrm{solution}\mathrm{is}$$$$\mathrm{x}=\{\frac{\pi }{8}+\mathrm{k}.\pi ;\frac{3\pi }{8}+\ell .\pi \}$$$$\mathrm{y}=\{-\frac{\pi }{8}+\mathrm{k}.\pi ;\frac{5\pi }{8}+\ell .\pi \}$$

Answered by MM42 last updated on 17/Sep/23

$$cos2x-cos2(x+y)=cos2x\Rightarrow 2(x+y)=k\pi +\frac{\pi }{2}$$$$2{cos}^2(x+y)-1=0\Rightarrow cos(x+y)=\pm \frac{\sqrt{2}}{2}$$$$\Rightarrow x+y=k\pi +\frac{\pi }{4};...,-\frac{3\pi }{4},\frac{\pi }{4},\frac{5\pi }{4},...$$$$2sin(x-y)cos(x+y)=\sqrt{2}$$$$\Rightarrow sin(x-y)=\pm 1$$$$\Rightarrow x-y=k\pi +\frac{\pi }{2};-\frac{\pi }{2},\frac{\pi }{2},\frac{3\pi }{2},...$$$$\Rightarrow x=k\pi +\frac{3\pi }{8}\mathrm{\&}y=k\pi -\frac{\pi }{8}$$$$$$

Answered by witcher3 last updated on 17/Sep/23

$$2\sin (2\mathrm{x}+\mathrm{y})\sin \left(\mathrm{y}\right)$$$$\cos \left(\mathrm{a}\right)-\cos \left(\mathrm{b}\right)=-2\sin \left(\frac{\mathrm{a}-\mathrm{b}}{2}\right)\sin \left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)$$$$\Rightarrow 2\mathrm{sn}(2\mathrm{x}+\mathrm{y})\sin \left(\mathrm{y}\right)=\cos \left(2\mathrm{x}\right)-\cos (2\mathrm{x}+2\mathrm{y})=\cos \left(2\mathrm{x}\right)$$$$\Rightarrow \cos (2x+2y)=0\Rightarrow 2\mathrm{x}+2\mathrm{y}=\frac{\pi }{2}+\mathrm{n}\pi$$$$\sin \left(2\mathrm{x}\right)-\sin \left(2\mathrm{y}\right)=\sqrt{2}$$$$\Rightarrow \sin (\frac{\pi }{2}+\mathrm{n}\pi -2\mathrm{y})-\sin \left(2\mathrm{y}\right)=\sqrt{2}$$$$\Rightarrow \cos (\mathrm{n}\pi -2\mathrm{y})-\sin \left(2\mathrm{y}\right)=\sqrt{2}$$$$\Rightarrow {(-1)}^{\mathrm{n}}\cos \left(2\mathrm{y}\right)-\sin \left(2\mathrm{y}\right)=\sqrt{2}$$$$\mathrm{n}=2\mathrm{k}\Rightarrow$$$$\cos \left(2\mathrm{y}\right)-\sin \left(2\mathrm{y}\right)=\sqrt{2}$$$$\Rightarrow \cos (2\mathrm{y}+\frac{\pi }{4})=1$$$$\Rightarrow 2\mathrm{y}+\frac{\pi }{4}=2\mathrm{k}\pi \Rightarrow \mathrm{y}=-\frac{\pi }{8}+\mathrm{m}\pi ,\mathrm{x}=\frac{3\pi }{8}+\mathrm{k}\pi$$$$\mathrm{n}=2\mathrm{k}+1$$$$2\mathrm{y}-\frac{\pi }{4}=\pi +2\mathrm{k}\pi$$$$\Rightarrow \mathrm{y}=\frac{5\pi }{8}+\mathrm{k}\pi$$$$\mathrm{x}=-\frac{\pi }{8}+\mathrm{n}\pi$$

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