What is the value of x if sinx=2 ?

What is the value of X if SinX =2

Obviously, there's no real x$x$ which satisfies the given condition.

We have an identity: sin(z)=eiz−e−iz2i$sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$

sin(x)=2$sin(x)=2$

=>eix−e−ix2i=2$=>\frac{e^{ix}-e^{-ix}}{2i}=2$

=>eix−e−ix=4i$=>e^{ix}-e^{-ix}=4i$

=>e2ix−4ieix−1=0,[eix≠0]$=>e^{2ix}-4i{e}^{ix}-1=0,[e^{ix}\ne 0]$

Let eix=t$e^{ix}=t$

So we have t2−4it−1=0$t^2-4it-1=0$

Using the quadratic formula to solve it, we get

x=$x=$2i±i√3=i(2±√3)$2i\pm i\surd 3=i(2\pm \surd 3)$

eix=i(2±√3)$e^{ix}=i(2\pm \surd 3)$

Taking natural log of both sides, we get

ix=ln(i(2±√3))$ix=ln(i(2\pm \surd 3))$

=>ix=ln(i)+ln(2±√3)$=>ix=ln(i)+ln(2\pm \surd 3)$

=>ix=iπ2+ln(2±√3)$=>ix=i\frac{\pi }{2}+ln(2\pm \surd 3)$

Dividing both sides by i$i$, we get

x=π2−iln(2±√3)$x=\frac{\pi }{2}-iln(2\pm \surd 3)$

Note that (2−√3)=12+√3$(2-\surd 3)=\frac{1}{2+\surd 3}$

So, x=π2−iln(2+√3)$x=\frac{\pi }{2}-iln(2+\surd 3)$ or π2−iln(12+√3)$\frac{\pi }{2}-iln(\frac{1}{2+\surd 3})$

=>x=π2−iln(2+√3)$=>x=\frac{\pi }{2}-iln(2+\surd 3)$ or π2+iln(2+√3)$\frac{\pi }{2}+iln(2+\surd 3)$

=>x=π2±iln(2+√3)$=>x=\frac{\pi }{2}\pm iln(2+\surd 3)$

We should also note that eiz+2nπ=eiz$e^{iz+2n\pi }=e^{iz}$,where n is an integer

Therefore x=π2±iln(2+√3)+2nπ$x=\frac{\pi }{2}\pm iln(2+\surd 3)+2n\pi$, n$n$belongs to integer.