Official solution
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 \[ = {{\sqrt {145} } \over {12}}\]