Question Number 153568 by liberty last updated on 08/Sep/21
$${solve}\:{in}\:{x}\in\mathbb{C} \\ $$$$\:\mathrm{sin}\:{x}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$ \\ $$
Answered by Ar Brandon last updated on 08/Sep/21
![sinx=(3/2) ⇒e^(ix) −e^(−ix) =3i ⇒e^(2ix) −3ie^(ix) −1=0 ⇒e^(ix) =((3i±i(√5))/2) ⇒ix=ln(i)+ln(3±(√5))−ln2 ⇒x=i[ln2−ln(3±(√5))−(π/2)i] =(π/2)+iln((2/(3±(√5))))=(π/2)−iln(((3±(√5))/2))](https://www.tinkutara.com/question/Q153570.png)
$$\mathrm{sin}{x}=\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow{e}^{{ix}} −{e}^{−{ix}} =\mathrm{3}{i} \\ $$$$\Rightarrow{e}^{\mathrm{2}{ix}} −\mathrm{3}{ie}^{{ix}} −\mathrm{1}=\mathrm{0}\:\Rightarrow{e}^{{ix}} =\frac{\mathrm{3}{i}\pm{i}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\Rightarrow{ix}=\mathrm{ln}\left({i}\right)+\mathrm{ln}\left(\mathrm{3}\pm\sqrt{\mathrm{5}}\right)−\mathrm{ln2} \\ $$$$\Rightarrow{x}={i}\left[\mathrm{ln2}−\mathrm{ln}\left(\mathrm{3}\pm\sqrt{\mathrm{5}}\right)−\frac{\pi}{\mathrm{2}}{i}\right] \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\pi}{\mathrm{2}}+{i}\mathrm{ln}\left(\frac{\mathrm{2}}{\mathrm{3}\pm\sqrt{\mathrm{5}}}\right)=\frac{\pi}{\mathrm{2}}−{i}\mathrm{ln}\left(\frac{\mathrm{3}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}\right) \\ $$