To find the total number of real solutions for the given equation, let's break down the problem step by step:

Define $\alpha$ as:

$ \alpha = \frac{1}{2} \sin^{-1} \left( \frac{6 \tan \theta}{9 + \tan^2 \theta} \right) $

We need to solve:

$ \theta = \tan^{-1}(2 \tan \theta) - \alpha $

This can be rearranged as:

$ \theta + \alpha = \tan^{-1}(2 \tan \theta) $

Taking the tangent of both sides, we have:

$ \tan(\theta + \alpha) = 2 \tan \theta $

The tangent addition formula gives:

$ \frac{\tan \theta + \tan \alpha}{1 - \tan \alpha \tan \theta} = 2 \tan \theta \quad \ldots (1) $

Next, consider:

$ \sin 2\alpha = \frac{6 \tan \theta}{9 + \tan^2 \theta} = \frac{2 \tan \alpha}{1 + \tan^2 \alpha} $

This leads to the equation:

$ 3 \tan \theta + 3 \tan \theta \tan^2 \alpha = 9 \tan \alpha + \tan \alpha \tan^2\theta $

Simplifying:

$ 3(\tan \theta - 3 \tan \alpha) = \tan \alpha \tan \theta (\tan \theta - 3 \tan \alpha) $

This implies:

$ \tan \theta = \frac{3}{\tan \alpha} \quad \text{or} \quad \tan \theta = 3 \tan \alpha $

Case I: $\tan \theta = 3 \tan \alpha$

From equation (1):

$ \frac{\tan \theta + \frac{\tan \theta}{3}}{1 - \frac{\tan^2 \theta}{3}} = 2 \tan \theta $

This gives:

$ \tan \theta = 0, \quad \frac{2}{3} = 1 - \frac{\tan^2 \theta}{3} \Rightarrow \tan \theta = 1, -1 $

Thus, we have:

$ \tan \theta = 0, -1, 1 \Rightarrow \theta = \frac{\pi}{4}, -\frac{\pi}{4}, 0 $

Case II: $\tan \theta = \frac{3}{\tan \alpha}$

From equation (1):

$ \frac{\tan \theta + \frac{3}{\tan \theta}}{-2} = 2 \tan \theta $

Simplifying this results in:

$ \tan \theta + \frac{3}{\tan \theta} = -4 \tan \theta $

This leads to no viable solutions (as denoted by "No Solution").

So, the real solutions for $\theta$ are $\frac{\pi}{4}$, $-\frac{\pi}{4}$, and $0$, resulting in a total of 3 real solutions.