The exhaustive solution set satisfying the equation  $3{\cos }^{-1}x={\sin }^{-1}\left(\left(\sqrt{1-x^2}\right)(4x^2-1)\right)$  is  [a,b] then [2a+b] equal to (where [.] denotes G.I.F).

$$\begin{gathered} 3{\cos }^{-1}x={\sin }^{-1}\left(\left(\sqrt{1-x^2}\right)(4x^2-1)\right) \\ x=\cos \theta , \theta \in [0,\pi ] \\ 3{\cos }^{-1}\left(\cos \theta \right)={\sin }^{-1}\left(\sin \theta (4{\cos }^2\theta -1)\right) \\ ={\sin }^{-1}(3\sin \theta -4{\sin }^3\theta ) ={\sin }^{-1}\left(\sin 3\theta \right) \end{gathered}$$

      Equality holds good is  $0\le \theta \le \pi$

$$\begin{gathered} -\frac{\pi }{2}\le 3\theta \le \frac{\pi }{2} \therefore 0\le \theta \le \frac{\pi }{6} \\ \frac{\sqrt{3}}{2}\le \cos \theta \le 1 \\ x\in [\frac{\sqrt{3}}{2},1] \end{gathered}$$