Given, ${\cos ^{ - 1}}\left( {{2 \over {3x}}} \right) + {\cos ^{ - 1}}\left( {{3 \over {4x}}} \right) = {\pi \over 2}$ $\Rightarrow {\cos ^{ - 1}}\left( {{2 \over {3x}}} \right) = {\pi \over 2} - {\cos ^{ - 1}}\left( {{3 \over {4x}}} \right)$ $\Rightarrow \cos \left( {{{\cos }^{ - 1}}\left( {{2 \over {3x}}} \right)} \right) = \cos \left[ {{\pi \over 2} - {{\cos }^{ - 1}}\left( {{3 \over {4x}}} \right)} \right]$ $\Rightarrow {2 \over {3x}} = \sin \left\{ {{{\cos }^{ - 1}}\left( {{3 \over {4x}}} \right)} \right\}$ $\Rightarrow {2 \over {3x}} = \sin \left\{ {{{\sin }^{ - 1}}{{\sqrt {16{x^2} - 9} } \over {4x}}} \right\}$ $\Rightarrow {2 \over {3x}} = {{16{x^2} - 9} \over {4x}}$ $\Rightarrow 64 = 9\left( {16{x^2} - 9} \right)$ $\Rightarrow 16{x^2} - 9 = {{64} \over 9}$ $\Rightarrow 16{x^2} = {{64} \over 9} + 9$ $\Rightarrow 16{x^2} = - {{145} \over 9}$ $\Rightarrow x = \pm {{\sqrt {145} } \over {4 \times 3}}$ $\Rightarrow x = \pm {{\sqrt {145} } \over {12}}$ as given that $x > {3 \over 4}$ $\therefore$ x $\ne$ $- {{\sqrt {145} } \over {12}}$ $\therefore$ x $= {{\sqrt {145} } \over {12}}$