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Question Number 121522 by benjo_mathlover last updated on 09/Nov/20

$$19\sin 2\mathrm{x}=37\cos 2\mathrm{x}+38\sin {}^2\mathrm{x}$$$$\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\tan \mathrm{x}.$$

Commented by liberty last updated on 09/Nov/20

$$\frac{38\sin \mathrm{xcos}\mathrm{x}}{\cos {}^2\mathrm{x}}=\frac{37(1-2\sin {}^2\mathrm{x})}{\cos {}^2\mathrm{x}}+38\tan {}^2\mathrm{x}$$$$\iff 38\tan \mathrm{x}=37(\sec {}^2\mathrm{x}-2\tan {}^2\mathrm{x})+38\tan {}^2\mathrm{x}$$$$\iff 38\tan \mathrm{x}=37-37\tan {}^2\mathrm{x}+38\tan {}^2\mathrm{x}$$$$\iff \tan {}^2\mathrm{x}-38\tan \mathrm{x}+37=0$$$$\Rightarrow (\tan \mathrm{x}-37)(\tan \mathrm{x}-1)=0$$$$\Rightarrow \tan \mathrm{x}=37\mathrm{or}\tan \mathrm{x}=1$$

Commented by MJS_new last updated on 09/Nov/20

$$\mathrm{yes}.$$$$\mathrm{you}\mathrm{could}\mathrm{simply}\mathrm{use}\mathrm{these},{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{usually}\mathrm{faster}$$$$(\mathrm{using}\sin x=s;\cos x=c;\tan x=t)$$$$s=\sqrt{1-c^2}=\frac{t}{\sqrt{t^2+1}}$$$$c=\sqrt{1-s^2}=\frac{1}{\sqrt{t^2+1}}$$$$t=\frac{s}{\sqrt{1-s^2}}=\frac{\sqrt{1-c^2}}{c}$$$$\sin 2x=\frac{2t}{t^2+1}$$$$\cos 2x=-\frac{t^2-1}{t^2+1}$$$$\tan 2x=-\frac{2t}{t^2-1}$$

Commented by benjo_mathlover last updated on 09/Nov/20

$$\mathrm{thank}\mathrm{you}$$

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