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Solve-sin-2x-2-




Question Number 153593 by tabata last updated on 08/Sep/21
Solve : (sin(2x))! = 2
$$\boldsymbol{{Solve}}\::\:\left(\boldsymbol{{sin}}\left(\mathrm{2}\boldsymbol{{x}}\right)\right)!\:=\:\mathrm{2} \\ $$
Answered by puissant last updated on 08/Sep/21
(sin(2x))!=2! ⇒ sin(2x)=2  ⇒ ((e^(2ix) −e^(−2ix) )/(2i))=2  ⇒ e^(4ix) −1=4ie^(2ix)   ⇒ e^(i4x) −4ie^(2ix) −1=0  z=e^(i2x)  , we have z^2 −4iz−1=0..  ...........
$$\left({sin}\left(\mathrm{2}{x}\right)\right)!=\mathrm{2}!\:\Rightarrow\:{sin}\left(\mathrm{2}{x}\right)=\mathrm{2} \\ $$$$\Rightarrow\:\frac{{e}^{\mathrm{2}{ix}} −{e}^{−\mathrm{2}{ix}} }{\mathrm{2}{i}}=\mathrm{2} \\ $$$$\Rightarrow\:{e}^{\mathrm{4}{ix}} −\mathrm{1}=\mathrm{4}{ie}^{\mathrm{2}{ix}} \\ $$$$\Rightarrow\:{e}^{{i}\mathrm{4}{x}} −\mathrm{4}{ie}^{\mathrm{2}{ix}} −\mathrm{1}=\mathrm{0} \\ $$$${z}={e}^{{i}\mathrm{2}{x}} \:,\:{we}\:{have}\:{z}^{\mathrm{2}} −\mathrm{4}{iz}−\mathrm{1}=\mathrm{0}.. \\ $$$$……….. \\ $$

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