Solve the equation $2\sin{ix} -3i\cos{ix}=3i$

complex numberscomplex-analysistrigonometry

I have got an answer for this, but I know that it is wrong because it contains $\cos^{-1}\left(-\frac{3}{\sqrt{5}}\right)$, which is undefined. So I was wondering if the issue is with the method I am using or if I have made a mistake with the numbers.

I rewrote the equation as
$$3i\cosh{x}-2i\sinh{x}=-3i$$
by using $\cos{ix}=\cosh{x}$ and $\sin{ix}=i\sinh{x}$. I then compared this to the compound angle formula for $\cos{a+bi}$, which is:
$$r\cos(a+bi)=r\cos{a}\cosh{b}-ir\sin{a}\sinh{b}$$
So, letting $r\cos{a}=3i$ and $r\sin{a}=2$ gives me $a=\tan^{-1}\frac{2}{3i}$ and $r=i\sqrt{5}$. This means that my original equation becomes:
$$\cos\left(\tan^{-1}\left(-\frac{2}{3}i\right)+ix\right)=-\frac{3}{\sqrt{5}}$$ which is impossible because the $\cos$ function will only produce values between $-1$ and $1$.

I can't see anything that is wrong with what I have done so I was wondering if I am missing something.

Best Answer

Hints:

  • Divide the equation you found by $i$.
  • Use the definition of $\cosh x = \dfrac {e^x + e^{-x}}{2}$ and $\sinh x = \dfrac {e^x - e^{-x}}{2}$ and simplify that equation.
  • Multiply the equation you simplified by $e^x$ to get rid of $e^{-x}$ and in turn get a quadratic in $e^x$.
  • Rearrange terms and factor; solve for $x$.

The solutions I got were $$x = i \pi + 2 \pi ik \\ x = -\ln 5 + i \pi + 2 \pi i k$$

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