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Question Number 116832 by bemath last updated on 07/Oct/20

$$\mathrm{If}19\sin 2\mathrm{x}=37\cos 2\mathrm{x}+38\sin {}^2\mathrm{x}$$$$\mathrm{then}\tan \mathrm{x}=\mathrm{\_}\mathrm{\_}$$

Answered by bobhans last updated on 07/Oct/20

$$\Rightarrow 19\sin 2\mathrm{x}=37\cos 2\mathrm{x}+38\sin {}^2\mathrm{x}$$$$\Rightarrow 19\sin 2\mathrm{x}-38\sin {}^2\mathrm{x}=37(\cos {}^2\mathrm{x}-\sin {}^2\mathrm{x})$$$$\Rightarrow 38\sin \mathrm{xcos}\mathrm{x}-38\sin {}^2\mathrm{x}=37(\cos \mathrm{x}-\sin \mathrm{x})(\cos \mathrm{x}+\sin \mathrm{x})$$$$\Rightarrow 38\sin \mathrm{x}(\cos \mathrm{x}-\sin \mathrm{x})=37(\cos \mathrm{x}-\sin \mathrm{x})(\cos \mathrm{x}+\sin \mathrm{x})$$$$\Rightarrow (\cos \mathrm{x}-\sin \mathrm{x})(38\sin \mathrm{x}-(37\cos \mathrm{x}+37\sin \mathrm{x}))=0$$$$\Rightarrow (\cos \mathrm{x}-\sin \mathrm{x})(\sin \mathrm{x}-37\cos \mathrm{x})=0$$$$\{\begin{array}{l}\cos \mathrm{x}=\sin \mathrm{x}\Rightarrow \tan \mathrm{x}=1\\ \sin \mathrm{x}=37\cos \mathrm{x}\Rightarrow \tan \mathrm{x}=37\end{array}$$

Answered by MJS_new last updated on 07/Oct/20

$$t=\tan x\iff x=\arctan t$$$$\frac{38t}{t^2+1}=-\frac{37(t^2-1)}{t^2+1}+\frac{38t^2}{t^2+1}$$$$t^2-38t+37=0$$$$\Rightarrow t=1\vee t=37$$$$\Rightarrow \tan x=1\vee \tan x=37$$

Commented by bobhans last updated on 07/Oct/20

$$\mathrm{your}\mathrm{are}\mathrm{amazing}\mathrm{sir}.\mathrm{i}\mathrm{want}\mathrm{learn}\mathrm{from}\mathrm{you}$$

Answered by malwaan last updated on 07/Oct/20

$$19\left[2sinxcosx\right]=37[{cos}^{2}x-$$$${sin}^{2}x]+38{sin}^{2}x$$$$\Rightarrow 38sinxcosx=37{cos}^{2}x+{sin}^{2}x$$$$\therefore {sin}^{2}x-38sinxcosx+37{cos}^{2}x=0$$$$(sinx-cosx)(sinx-37cosx)=0$$$$sinx-cosx=0\Rightarrow tanx=1$$$$sinx-37cosx=0\Rightarrow tanx=37$$

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