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Author: Tinku Tara

Question-126712

Question Number 126712 by help last updated on 23/Dec/20 Commented by liberty last updated on 24/Dec/20 $$\:\left(\bullet\right)\:{x}+\frac{\mathrm{1}}{{x}}\:=\:{w}\:\Rightarrow{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:=\:{w}^{\mathrm{2}} −\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{2}\:=\:{w}^{\mathrm{2}} −\mathrm{2} \\…

solve-x-x-2-1-

Question Number 126708 by slahadjb last updated on 23/Dec/20 $$\boldsymbol{{solve}}\:\:\:\:\:\boldsymbol{{x}}+\boldsymbol{{x}}^{\sqrt{\mathrm{2}}} =\mathrm{1} \\ $$ Commented by Dwaipayan Shikari last updated on 23/Dec/20 $$\sim\mathrm{0}.\mathrm{559793} \\ $$ Commented…

Question-192241

Question Number 192241 by yaslm last updated on 12/May/23 Answered by Frix last updated on 12/May/23 $${Z}_{\mathrm{1}} =\frac{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{e}^{−\mathrm{i}{x}} }{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{e}^{−\mathrm{i}{x}} }=−\frac{\mathrm{1}−\mathrm{2cos}\:{x}}{\mathrm{5}−\mathrm{4cos}\:{x}}−\frac{\mathrm{2sin}\:{x}}{\mathrm{5}−\mathrm{4cos}\:{x}}\mathrm{i} \\ $$$${Z}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{e}^{{ix}} }=\frac{\mathrm{2}\left(\mathrm{2}−\mathrm{cos}\:{x}\right)}{\mathrm{5}−\mathrm{4cos}\:{x}}+\frac{\mathrm{2sin}\:{x}}{\mathrm{5}−\mathrm{4cos}\:{x}}\mathrm{i} \\…