Help solving (complex) simultaneous equations

[KFC]

Homework Statement

I am trying to solving the following complex equation for $$x$$ and $$\theta$$

$$a\sinh(2x) e^{-i\theta} + y\sinh^2x e^{-i2\theta} + y^*\cosh^2(x) = 0$$

where $$a$$ is real constant, $$x$$ and $$\theta$$ are also real parameter. $$y$$ is complex number, $$y^*$$ is the complex conjugate.

Solve for $$x$$ and $$\theta$$ (in terms of y and a)

2. The attempt at a solution
Let
$$y = |y| e^{i\varphi}$$

and multiply the equation with $$y$$

$$ay\sinh(2x) e^{-i\theta} + y^2\sinh^2x e^{-i2\theta} + |y|^2\cosh^2(x) = 0$$

Now let the real part and imaginary part equals ZERO.

$$\begin{cases}<br /> a\sinh(2x) |y|\cos(\theta-\varphi) + |y|^2\sinh^2(x)\cos(2\theta-2\varphi) + |y|^2\alpha^2 = 0, \\[3.8mm]<br /> a\sinh(2x) |y|\sin(\theta-\varphi) + |y|^2\sinh^2(x)\sin(2\theta-2\varphi) = 0<br /> \end{cases}$$

I tryied to solve that two days ago, I tried many way to simpliy that but still find no way to get the soluton. Could anyone give me some hints?

Thanks

 

[Mark44]

This doesn't make sense to me. How can you solve one equation for two variables? This seems to me like asking someone to solve y = 2x for x and y. You can solve the equation for y in terms of x, or you can solve for x in terms of y, but you can't solve it simultaneously for both variables.

 

[KFC]

This doesn't make sense to me. How can you solve one equation for two variables? This seems to me like asking someone to solve y = 2x for x and y. You can solve the equation for y in terms of x, or you can solve for x in terms of y, but you can't solve it simultaneously for both variables.

How come. This is a equation for complex variable, the real part and imaginary part gives two equations.

 

[Mark44]

OK, I see.
$$a\sinh(2x) |y|\sin(\theta-\varphi) + |y|^2\sinh^2(x)\sin(2\theta-2\varphi) = 0$$
In your 2nd equation, I think the sine arguments should be the other way around. Also, you can expand $sin(2(\phi - \theta))$ as $2sin(\phi - \theta) cos(\phi - \theta)$. Then you'll have one factor the same in both terms, which hopefully leaves you with the other factor that you can do something with.