Finding a solution to sin(x) = -5 using Euler's Formula

I have just learned about Euler's formula and I am attempting to find a solution to $\sin(\theta)=-5$. However, I am not entirely sure how to precede.

Thank You So Much!

• What do you mean? $-1 \le \sin \theta \le 1$ !! – SchrodingersCat Nov 22 '15 at 17:49
• But I through with the inclusion of complex numbers the domain of sin(x) is all real numbers? – sbmit2 Nov 22 '15 at 17:53
• @Aniket Euler formula is about complex numbers. – N. S. Nov 22 '15 at 17:54
• @N.S. I know Euler's formula. But OP had used $x$ without mentioning that he was considering it as complex number. He commented on it now only. – SchrodingersCat Nov 22 '15 at 17:57
• @Aniket He did put the complex number tag though. – user236182 Nov 22 '15 at 18:00

HINT (you only need to simplify the final answer):

$$\sin(x)=-5\Longleftrightarrow$$ $$\frac{1}{2}\left(ie^{-ix}-ie^{ix}\right)=-5\Longleftrightarrow$$ $$ie^{-ix}-ie^{ix}=-10\Longleftrightarrow$$

Substitute $y=-ie^{ix}$:

$$\frac{1}{(0-i)e^{(0+i)x}}+(0-i)e^{(0+i)x}=-10\Longleftrightarrow$$ $$y+\frac{1}{y}=-10\Longleftrightarrow$$ $$\frac{y^2+1}{y}=-10\Longleftrightarrow$$ $$y^2+1=-10\Longleftrightarrow$$ $$y^2+10y+1=0\Longleftrightarrow$$ $$y^2+10y=-1\Longleftrightarrow$$ $$y^2+10y+25=24\Longleftrightarrow$$ $$(y+5)^2=24\Longleftrightarrow$$ $$y+5=\pm\sqrt{24}\Longleftrightarrow$$ $$y+5=\pm 2\sqrt{6}\Longleftrightarrow$$ $$y=\pm 2\sqrt{6}-5\Longleftrightarrow$$ $$-ie^{ix}=\pm 2\sqrt{6}-5\Longleftrightarrow$$ $$e^{ix}=\frac{\pm 2\sqrt{6}-5}{-i}\Longleftrightarrow$$ $$ix=\ln\left(\frac{\pm 2\sqrt{6}-5}{-i}\right)+2i\pi n\Longleftrightarrow$$ $$x=\frac{\ln\left(\frac{\pm 2\sqrt{6}-5}{-i}\right)+2i\pi n}{i}$$

$$e^{i\theta}-e^{-i\theta}=-10i\\ z-z^{-1}=-10i\\ z^2-1=-10iz$$ Solve the quadratic, then take logarithm to find $\theta$.

• or take a logarithm,as you have many to choose from – DanielWainfleet Nov 22 '15 at 18:13