Question:

<div class="solution"> <p><strong>Step 1:</strong> We are given the equation \(\alpha = \frac{1}{3}\sin^{-1}\left(\frac{2x}{1+x^2}\right) + \frac{1}{3}\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\) where \(x \geq \frac{4}{3}\). To simplify this, let's consider the expressions inside the inverse trigonometric functions. Recall that \(\sin^{-1}\left(\frac{2x}{1+x^2}\right)\) and \(\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\) can be related to the tangent of a half-angle formula, specifically \(\tan\left(\frac{\theta}{2}\right) = \frac{x}{1}\), which implies \(x = \tan\left(\frac{\theta}{2}\right)\). Using the double angle formulas, we can express \(\sin(\theta)\) and \(\cos(\theta)\) in terms of \(x\), which are \(\sin(\theta) = \frac{2x}{1+x^2}\) and \(\cos(\theta) = \frac{1-x^2}{1+x^2}\).</p> <p><strong>Step 2:</strong> To evaluate \(\alpha\), we recognize that \(\sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\sin^{-1}(x)\) and \(\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = 2\cos^{-1}(x)\) for \(x \geq \frac{4}{3}\) may not directly apply due to the range of \(x\). However, considering the relationship of \(\alpha\) with \(x\), and knowing that \(\sin^{-1}\) and \(\cos^{-1}\) are related to the angles whose sine and cosine are the given values, we aim to express \(\alpha\) in terms of \(x\) and known angles to facilitate the computation of \(\cos 2\alpha\), \(\sec \alpha\), and the overall expression \(\dfrac{\cos 2\alpha + \sec\alpha + 3\sqrt{3}}{\sqrt{3}}\). Given the constraints and aiming for simplification, we should look into trigonometric identities that relate \(\sin^{-1}\), \(\cos^{-1}\), and the given expression for \(\alpha\).</p> <p><strong>Step 3:</strong> Let's analyze the given options and the structure of the problem to deduce a logical path. The expression involves \(\cos 2\alpha\) and \(\sec \alpha\), suggesting the use of double angle and reciprocal identities. Considering \(\alpha = \frac{1}{3}\sin^{-1}\left(\frac{2x}{1+x^2}\right) + \frac{1}{3}\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\), and knowing that \(x \geq \frac{4}{3}\), we should examine the behavior of \(\sin^{-1}\) and \(\cos^{-1}\) functions within this domain and their impact on \(\alpha\). Utilizing the identity \(\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}\) for \(x \in [-1,1]\), we recognize the need to adjust our approach since our \(x\) is outside this range but the arguments of \(\sin^{-1}\) and \(\cos^{-1}\) in \(\alpha\) are within \([-1,1]\).</p> <p><strong>Step 4:</strong> To find \(\dfrac{\cos 2\alpha + \sec\alpha + 3\sqrt{3}}{\sqrt{3}}\), let's first simplify \(\alpha\). Given \(\alpha = \frac{1}{3}\sin^{-1}\left(\frac{2x}{1+x^2}\right) + \frac{1}{3}\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\), and recognizing the relationship between \(\sin^{-1}\) and \(\cos^{-1}\), consider that \(\sin^{-1}\left(\frac{2x}{1+x^2}\right) + \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = \frac{\pi}{2}\) because the sum of angles whose sine and cosine yield specific values