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

$$2\sin{ix} -3i\cos{ix}=3i<=>$$
$$ <=>\sqrt{4+9i^2} \sin \left(ix -\arcsin {\frac{3i}{\sqrt{4+9i^2}}}\right)=3i<=>$$
$$ <=>i \sqrt{5} \sin \left(ix -\arcsin {\frac{3}{\sqrt{5}}}\right)=3i<=>$$
$$ <=>\sin \left(ix -\arcsin {\frac{3}{\sqrt{5}}}\right)=\frac{3} {\sqrt{5}}<=>$$
$$ <=>\left[\begin{array} {1} ix -\arcsin {\frac{3}{\sqrt{5}}}=\arcsin {\frac{3}{\sqrt{5}}} + 2\pi n\\ ix -\arcsin {\frac{3}{\sqrt{5}}}=\pi-\arcsin {\frac{3}{\sqrt{5}}}+ 2\pi n \end{array} \right. <=>$$
$$ <=>\left[\begin{array} {1} ix =2\arcsin {\frac{3}{\sqrt{5}}} + 2\pi n\\ ix = \pi+ 2\pi n \end{array} \right. <=>$$
$$ <=>\left[\begin{array} {1} ix = \pi - i \ln{5} + 2\pi n\\ ix = \pi+ 2\pi n \end{array} \right. <=>$$
$$ <=>\left[\begin{array} {1} x = \pi i + \ln{5} + 2\pi n i\\ x = \pi i+ 2\pi n i \end{array} \right. $$
