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Question Number 45280 by peter frank last updated on 11/Oct/18

$$\mathbf{solve}\mathbf{the}\mathbf{simultaneous}\mathbf{equation}$$$$\mathbf{a})\mathbf{sin}(\mathbf{x}+\mathbf{y})=\frac{1}{\sqrt{2}}$$$$\mathbf{cos}2\mathbf{x}=\frac{-1}{2}\mathbf{for}\mathbf{x}\mathbf{and}\mathbf{y}\mathbf{ranging}\mathbf{from}0\mathbf{to}360\mathbf{inclusive}$$$$\mathbf{b})\mathbf{if}\mathbf{siny}+\mathbf{cosx}=\mathbf{x}\mathbf{show}\mathbf{that}\frac{{\mathbf{d}}^2\mathbf{y}}{{\mathbf{dx}}^2}=\frac{\mathbf{x}}{{(2-{\mathbf{x}}^2)}^{\frac{3}{2}}}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18

$$sin(x+y)=\frac{1}{\sqrt{2}}=sin(\pi -\frac{\pi }{4})$$$$x+y=\frac{3\pi }{4}$$$$cos2x=-\frac{1}{2}=cos(\pi -\frac{\pi }{3})$$$$2x=\frac{2\pi }{3}x=\frac{\pi }{3}y=\frac{3\pi }{4}-\frac{\pi }{3}=\frac{5\pi }{12}$$$$x=\frac{\pi }{3}y=\frac{5\pi }{12}$$$$$$$$$$$$$$

Commented by peter frank last updated on 11/Oct/18

$$\mathrm{thank}\mathrm{you}\mathrm{sir}.\mathrm{pls}\mathrm{help}\mathrm{qn}2$$

Answered by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

$$\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{x}{{(2-x^2)}^{\frac{3}{2}}}$$$$\int d\left(\frac{dy}{dx}\right)=\int \frac{xdx}{{(2-x^2)}^{\frac{3}{2}}}$$$$t^2=2-x^{2}2tdt=-2xdx$$$$RHS=\int \frac{-tdt}{t^3}$$$$=\int -t^{-2}dt$$$$=-1\times \frac{t^{-1}}{-1}=\frac{1}{t}=\frac{1}{\sqrt{2-x^2}}$$$$\frac{dy}{dx}=\frac{1}{\sqrt{2-x^2}}$$$$\int dy=\int \frac{dx}{\sqrt{2-x^2}}x=\sqrt{2}sin\theta dx=\sqrt{2}cos\theta d\theta$$$$y=\int \frac{\sqrt{2}cos\theta }{\sqrt{2}cos\theta }d\theta$$$$y=\theta +c$$$$y={sin}^{-1}\left(\frac{x}{\sqrt{2}}\right)+c$$$$sin(y-c)=\frac{x}{\sqrt{2}}$$$$$$$$x=\sqrt{2}sin(y-c)$$$$SOplscheckthequestion....$$

Commented by peter frank last updated on 12/Oct/18

$$\mathrm{its}\mathrm{seem}\mathrm{there}\mathrm{are}\mathrm{something}\mathrm{wrong}.\mathrm{thanks}\mathrm{sir}$$$$$$

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