The total number of real solutions of the equation $\theta = \tan^{-1}(2\tan\theta) - \frac{1}{2}\sin^{-1}\left(\frac{6\tan\theta}{9+\tan^{2}\theta}\right)$ is (Here, the inverse trigonometric functions $\sin^{-1}x$ and $\tan^{-1}x$ assume values in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ and $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, respectively.)

$\theta = \tan^{-1}(2\tan\theta) - \dfrac{1}{2}\sin^{-1}\left(\dfrac{6\tan\theta}{9+\tan^2\theta}\right)$ $\sin^{-1} \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right], \quad \tan^{-1} \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ $\text{To find: total number of real solutions } \theta$ $t = \tan\theta, \quad \alpha = \tan^{-1}\!\left(\dfrac{t}{3}\right)$ $\sin 2\alpha = \dfrac{2(t/3)}{1 + t^2/9} = \dfrac{6t}{9 + t^2}$ $\tan^{-1} \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right), \quad \dfrac{1}{2}\sin^{-1} \in \left[-\dfrac{\pi}{4}, \dfrac{\pi}{4}\right]$ $\Rightarrow \theta \in \left(-\dfrac{3\pi}{4}, \dfrac{3\pi}{4}\right), \ \theta \neq \pm\dfrac{\pi}{2}$ $\theta \in \left(\dfrac{\pi}{2}, \dfrac{3\pi}{4}\right): \text{RHS} < \dfrac{\pi}{4} < \theta \Rightarrow \text{no solution}$ $(\text{and symmetrically for } \theta \in (-\tfrac{3\pi}{4}, -\tfrac{\pi}{2}))$ $|t| \le 3:\ \sin^{-1}(\sin 2\alpha) = 2\alpha \Rightarrow \dfrac{1}{2}\sin^{-1}(\cdots) = \alpha = \tan^{-1}\dfrac{t}{3}$ $\theta = \tan^{-1}(2t) - \tan^{-1}\dfrac{t}{3}$ $\tan^{-1}(2t) - \tan^{-1}\dfrac{t}{3} = \tan^{-1}\dfrac{2t - t/3}{1 + 2t\cdot t/3}$ $= \tan^{-1}\dfrac{5t}{3 + 2t^2}$ $\theta = \tan^{-1}\dfrac{5t}{3 + 2t^2} \Rightarrow \tan\theta = \dfrac{5t}{3 + 2t^2}$ $t = \dfrac{5t}{3 + 2t^2}$ $t(3 + 2t^2) = 5t \Rightarrow 2t(t^2 - 1) = 0$ $t = 0, \ t = \pm 1$ $t = \tan\theta \Rightarrow \theta = 0, \ \dfrac{\pi}{4}, \ -\dfrac{\pi}{4}$ $t > 3:\ \dfrac{1}{2}\sin^{-1}(\sin 2\alpha) = \dfrac{\pi}{2} - \alpha$ $\text{RHS} = \tan^{-1}(2t) + \tan^{-1}\dfrac{t}{3} - \dfrac{\pi}{2} < \theta \Rightarrow \text{no solution}$ $\boxed{\text{3 real solutions}} \quad \text{option (3)}$