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.)

Let $$t=\tan\theta$$. The given equation becomes

$$\theta=\tan^{-1}(2t)-\frac12\sin^{-1}\!\left(\frac{6t}{9+t^{2}}\right)\qquad -(1)$$

Step 1 : Simplify the $$\sin^{-1}$$ term.
Put $$t=3u$$ so that $$u=\dfrac{t}{3}$$. Then

$$\frac{6t}{9+t^{2}}=\frac{6\,(3u)}{9+9u^{2}} =\frac{18u}{9(1+u^{2})} =\frac{2u}{1+u^{2}} =\sin 2\alpha,$$
where $$\alpha=\tan^{-1}u=\tan^{-1}\!\left(\dfrac{t}{3}\right).$$

The principal value of $$\sin^{-1}(\sin 2\alpha)$$ equals $$2\alpha$$ whenever $$|2\alpha|\le\dfrac{\pi}{2}$$, i.e. $$|\alpha|\le\dfrac{\pi}{4}$$. Because $$|\alpha|\le\dfrac{\pi}{4}\iff |u|\le1\iff |t|\le3,$$ the replacement is valid for every $$t$$ that will survive the later algebra (it will turn out that $$|t|\le1$$).

Hence

$$\sin^{-1}\!\left(\frac{6t}{9+t^{2}}\right)=2\alpha =2\tan^{-1}\!\left(\frac{t}{3}\right).$$

Step 2 : Substitute back in equation $$(1)$$.

$$\theta=\tan^{-1}(2t)-\tan^{-1}\!\left(\frac{t}{3}\right)\qquad -(2)$$

Step 3 : Use the difference formula for $$\tan^{-1}$$.
For real $$a,b$$ within the principal interval,

$$\tan^{-1}a-\tan^{-1}b=\tan^{-1}\!\left(\frac{a-b}{1+ab}\right).$$

Applying this with $$a=2t,\; b=\dfrac{t}{3}$$ gives

$$\theta=\tan^{-1}\!\left(\frac{2t-\dfrac{t}{3}} {1+\dfrac{2t^{2}}{3}}\right) =\tan^{-1}\!\left(\frac{5t}{3+2t^{2}}\right)\qquad -(3)$$

Step 4 : Equate tangents.
Take tangent on both sides of $$(3)$$ (possible because both sides now lie in $$\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$$):

$$t=\frac{5t}{3+2t^{2}}.$$

Multiply by $$3+2t^{2}$$:

$$t(3+2t^{2})=5t\quad\Longrightarrow\quad 3t+2t^{3}=5t\quad\Longrightarrow\quad 2t^{3}-2t=0.$$

Thus

$$t\bigl(t^{2}-1\bigr)=0\quad\Longrightarrow\quad t=0,\; t=1,\; t=-1.$$

These values satisfy $$|t|\le1\le3$$, so the earlier replacement in Step 1 is justified.

Step 5 : Find the corresponding $$\theta$$ values.

Case 1: $$t=0\Rightarrow \tan\theta=0\Rightarrow \theta=0.$$ For any other integer multiple of $$\pi$$, the right side of the original equation is still $$0$$ while the left side is $$\pm\pi,\pm2\pi,\ldots$$, so only $$\theta=0$$ works. Case 2: $$t=1\Rightarrow \tan\theta=1\Rightarrow \theta=\frac{\pi}{4}+n\pi.$$ Using principal values $$\tan^{-1}(2)=1.107,\;\sin^{-1}\!\left(\frac35\right)=0.6435,$$ the right side of $$(1)$$ equals $$0.7854=\frac{\pi}{4}.$$ Hence $$\theta$$ must be $$\frac{\pi}{4}$$ (the choice $$n=0$$). Other choices of $$n$$ give $$\theta=\frac{5\pi}{4},\,-\frac{3\pi}{4},\ldots,$$ which lie outside the principal range of the inverse functions and therefore do not satisfy $$(1)$$. Case 3: $$t=-1\Rightarrow \tan\theta=-1\Rightarrow \theta=-\frac{\pi}{4}+n\pi.$$ Now the right side of $$(1)$$ equals $$-0.7854=-\frac{\pi}{4},$$ so only $$\theta=-\frac{\pi}{4}$$ (again $$n=0$$) works.

Step 6 : Count the solutions.
The equation is satisfied for exactly

$$\theta=-\frac{\pi}{4},\; 0,\; \frac{\pi}{4}.$$

Therefore the total number of real solutions is $$3$$.

Option C which is: 3