Question:

Step 1: Express trigonometric functions in terms of $\tan\alpha$. Let $t = \tan\alpha$. Using the double angle formulas in terms of $t$: $$\sin 2\alpha = \frac{2t}{1+t^2}, \quad \cos 2\alpha = \frac{1-t^2}{1+t^2}$$ Step 2: Simplify the argument of the inverse sine function. We need to simplify $\dfrac{3\sin 2\alpha}{5+4\cos 2\alpha}$. Substituting the expressions from Step 1: $$\frac{3\sin 2\alpha}{5+4\cos 2\alpha} = \frac{3 \cdot \frac{2t}{1+t^2}}{5+4 \cdot \frac{1-t^2}{1+t^2}}$$ Simplifying the denominator: $$5+4\cos 2\alpha = \frac{5(1+t^2)+4(1-t^2)}{1+t^2} = \frac{5+5t^2+4-4t^2}{1+t^2} = \frac{9+t^2}{1+t^2}$$ Therefore: $$\frac{3\sin 2\alpha}{5+4\cos 2\alpha} = \frac{\frac{6t}{1+t^2}}{\frac{9+t^2}{1+t^2}} = \frac{6t}{9+t^2}$$ Step 3: Recognize the expression as a sine of double angle. Notice that $\dfrac{6t}{9+t^2}$ can be written using the double angle formula for sine. Let $\beta = \tan^{-1}\left(\dfrac{t}{3}\right)$. Then: $$\sin 2\beta = \frac{2\tan\beta}{1+\tan^2\beta} = \frac{2 \cdot \frac{t}{3}}{1+\left(\frac{t}{3}\right)^2} = \frac{\frac{2t}{3}}{\frac{9+t^2}{9}} = \frac{6t}{9+t^2}$$ Therefore: $$\frac{6t}{9+t^2} = \sin\left(2\tan^{-1}\left(\frac{t}{3}\right)\right)$$ Step 4: Apply the inverse sine function. Taking the inverse sine of both sides (for the appropriate range): $$\sin^{-1}\left(\frac{6t}{9+t^2}\right) = 2\tan^{-1}\left(\frac{t}{3}\right)$$ Step 5: Solve for $x$. Substituting this into the original equation: $$\frac{1}{2}\sin^{-1}\left(\frac{3\sin 2\alpha}{5+4\cos 2\alpha}\right) = \frac{1}{2} \cdot 2\tan^{-1}\left(\frac{t}{3}\right) = \tan^{-1}\left(\frac{t}{3}\right)$$ Since this equals $\tan^{-1} x$: $$\tan^{-1} x = \tan^{-1}\left(\frac{t}{3}\right)$$ Therefore: $$x = \frac{t}{3} = \frac{\tan\alpha}{3}$$ **Final Answer:** The possible value of $x$ is $\dfrac{\tan\alpha}{3}$, which can be written as $\dfrac{1}{3}\tan\alpha$. The answer is **Option 3**.