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tan-x-4-cos-2x-cot-2x-




Question Number 103823 by bobhans last updated on 17/Jul/20
tan (x) = 4 cos (2x)−cot (2x)
$$\mathrm{tan}\:\left({x}\right)\:=\:\mathrm{4}\:\mathrm{cos}\:\left(\mathrm{2}{x}\right)−\mathrm{cot}\:\left(\mathrm{2}{x}\right) \\ $$
Answered by Dwaipayan Shikari last updated on 17/Jul/20
((sinx)/(cosx))=cos2x(4−(1/(sin2x)))  ((sinx)/(cosx))=((2sin4x−cos2x)/(2sinxcosx))  2sin^2 x=2sin4x−cos2x  1−cos2x=2sin4x−cos2x  sin4x=(1/2)  4x=2kπ+(−1)^k (π/6)  x=((kπ)/2)±(π/(24))=(12k±1)(π/(24))= { k,m∈Z   (general solution
$$\frac{{sinx}}{{cosx}}={cos}\mathrm{2}{x}\left(\mathrm{4}−\frac{\mathrm{1}}{{sin}\mathrm{2}{x}}\right) \\ $$$$\frac{{sinx}}{{cosx}}=\frac{\mathrm{2}{sin}\mathrm{4}{x}−{cos}\mathrm{2}{x}}{\mathrm{2}{sinxcosx}} \\ $$$$\mathrm{2}{sin}^{\mathrm{2}} {x}=\mathrm{2}{sin}\mathrm{4}{x}−{cos}\mathrm{2}{x} \\ $$$$\mathrm{1}−{cos}\mathrm{2}{x}=\mathrm{2}{sin}\mathrm{4}{x}−{cos}\mathrm{2}{x} \\ $$$${sin}\mathrm{4}{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{4}{x}=\mathrm{2}{k}\pi+\left(−\mathrm{1}\right)^{{k}} \frac{\pi}{\mathrm{6}} \\ $$$${x}=\frac{{k}\pi}{\mathrm{2}}\pm\frac{\pi}{\mathrm{24}}=\left(\mathrm{12}{k}\pm\mathrm{1}\right)\frac{\pi}{\mathrm{24}}=\:\left\{\:{k},{m}\in\mathbb{Z}\:\:\:\left({general}\:{solution}\right.\right. \\ $$
Answered by bobhans last updated on 17/Jul/20

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