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Question Number 58804 by Tawa1 last updated on 30/Apr/19

$$\mathrm{Find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{x}:4\cos \mathrm{x}+3\sin \mathrm{x}=2$$

Commented by MJS last updated on 30/Apr/19

$$\mathrm{generally}$$$$a\cos x+b\sin x=c$$$$x=2\arctan t$$$$t^2-\frac{2b}{a+c}t-\frac{a-c}{a+c}=0$$$$t=\frac{b\pm \sqrt{a^2+b^2-c^2}}{a+c}$$$$x=2\pi n+2\arctan \frac{b\pm \sqrt{a^2+b^2-c^2}}{a+c};n\in \mathbb{Z}$$$$\mathrm{even}\mathrm{if}t\notin \mathbb{R}$$

Answered by MJS last updated on 30/Apr/19

$$4\cos x+3\sin x=2$$$$x=2\arctan t$$$$4\frac{1-t^2}{1+t^2}+3\frac{2t}{1+t^2}=2$$$$t^2-t-\frac{1}{3}=0$$$$t=\frac{1}{2}\pm \frac{\sqrt{21}}{6}$$$$x=2\pi n+2\arctan \frac{3\pm \sqrt{21}}{6};n\in \mathbb{Z}$$

Commented by Tawa1 last updated on 30/Apr/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by Tawa1 last updated on 30/Apr/19

$$\mathrm{Sir},\mathrm{why}\mathrm{when}\mathrm{n}=1,\mathrm{the}\mathrm{equation}\mathrm{is}\mathrm{not}\mathrm{equal}\mathrm{to}2$$

Answered by mr W last updated on 30/Apr/19

$$4\cos \mathrm{x}+3\sin \mathrm{x}=2$$$$\frac{4}{5}\cos \mathrm{x}+\frac{3}{5}\sin \mathrm{x}=\frac{2}{5}$$$$\cos \alpha \cos \mathrm{x}+\sin \alpha \sin \mathrm{x}=\frac{2}{5}$$$$\cos (\mathrm{x}-\alpha )=\frac{2}{5}$$$$\Rightarrow x-\alpha =2n\pi \pm {\cos }^{-1}\frac{2}{5}$$$$\Rightarrow x=2n\pi +\alpha \pm {\cos }^{-1}\frac{2}{5}$$$$\Rightarrow x=2n\pi +{\cos }^{-1}\frac{4}{5}\pm {\cos }^{-1}\frac{2}{5}$$

Commented by Tawa1 last updated on 30/Apr/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by Tawa1 last updated on 30/Apr/19

$$\mathrm{Sir},\mathrm{when}\mathrm{n}=1,\mathrm{the}\mathrm{equation}\mathrm{is}\mathrm{not}\mathrm{equal}\mathrm{to}2.$$

Commented by mr W last updated on 30/Apr/19

$$howdidyoucometothat?$$$$$$$$withn=1:$$$$x=2\pi +{\cos }^{-1}\frac{4}{5}+{\cos }^{-1}\frac{2}{5}$$$$or$$$$x=2\pi +{\cos }^{-1}\frac{4}{5}-{\cos }^{-1}\frac{2}{5}$$$$$$$$withx=2\pi +{\cos }^{-1}\frac{4}{5}+{\cos }^{-1}\frac{2}{5}:$$$$\cos x=\cos ({\cos }^{-1}\frac{4}{5}+{\cos }^{-1}\frac{2}{5})$$$$=\frac{4}{5}\times \frac{2}{5}-\frac{3}{5}\times \frac{\sqrt{21}}{5}=\frac{8-3\sqrt{21}}{25}$$$$\sin x=\sin ({\cos }^{-1}\frac{4}{5}+{\cos }^{-1}\frac{2}{5})$$$$=\frac{3}{5}\times \frac{2}{5}+\frac{4}{5}\times \frac{\sqrt{21}}{5}=\frac{6+4\sqrt{21}}{25}$$$$4\cos x+3\sin x=\frac{4(8-3\sqrt{21})+3(6+4\sqrt{21})}{25}$$$$=\frac{32+18}{25}=\frac{50}{25}=2!$$$$$$$$withx=2\pi +{\cos }^{-1}\frac{4}{5}-{\cos }^{-1}\frac{2}{5}:$$$$\cos x=\cos ({\cos }^{-1}\frac{4}{5}-{\cos }^{-1}\frac{2}{5})$$$$=\frac{4}{5}\times \frac{2}{5}+\frac{3}{5}\times \frac{\sqrt{21}}{5}=\frac{8+3\sqrt{21}}{25}$$$$\sin x=\sin ({\cos }^{-1}\frac{4}{5}-{\cos }^{-1}\frac{2}{5})$$$$=\frac{3}{5}\times \frac{2}{5}-\frac{4}{5}\times \frac{\sqrt{21}}{5}=\frac{6-4\sqrt{21}}{25}$$$$4\cos x+3\sin x=\frac{4(8+3\sqrt{21})+3(6-4\sqrt{21})}{25}$$$$=\frac{32+18}{25}=\frac{50}{25}=2!$$

Commented by Tawa1 last updated on 30/Apr/19

$$\mathrm{Sir},\mathrm{when}\mathrm{i}\mathrm{used}\pi =360,\mathrm{i}\mathrm{got}2\mathrm{now}.\mathrm{Before}\mathrm{i}\mathrm{used}\pi =3.142$$

Commented by MJS last updated on 30/Apr/19

$$\mathrm{maybe}\mathrm{on}\mathrm{your}\mathrm{calculator}\mathrm{you}\mathrm{must}\mathrm{set}\mathrm{the}$$$$\mathrm{mode}\mathrm{for}\mathrm{the}\mathrm{angle}$$$$\mathrm{usually}{\mathrm{it}}^{\prime }\mathrm{s}‘‘{deg}^{″}\mathrm{for}\mathrm{degree}(\mathrm{full}\mathrm{circle}=360°)$$$$\mathrm{and}‘‘{rad}^{″}\mathrm{for}\mathrm{radiant}(\mathrm{full}\mathrm{circle}=2\pi )$$

Commented by Tawa1 last updated on 30/Apr/19

$$\mathrm{Ohh}.\mathrm{Thanks}\mathrm{for}\mathrm{your}\mathrm{time}\mathrm{sir}$$

Commented by MJS last updated on 30/Apr/19

$${\mathrm{you}}^{\prime }\mathrm{re}\mathrm{welcome}$$

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