If $${\cos ^{ - 1}}\left( {\frac{2}{{3x}}} \right) + {\cos ^{ - 1}}\left( {\frac{3}{{4x}}} \right) = \frac{\pi }{2}\left( {x > \frac{3}{4}} \right),$$         then

Solution :
$$\eqalign{ & {\cos ^{ - 1}}\left( {\frac{2}{{3x}}} \right) + {\cos ^{ - 1}}\left( {\frac{3}{{4x}}} \right) = \frac{\pi }{2};\left( {x > \frac{3}{4}} \right) \cr & \Rightarrow \,\,{\cos ^{ - 1}}\left( {\frac{2}{{3x}}} \right)\frac{\pi }{2} - {\cos ^{ - 1}}\left( {\frac{3}{{4x}}} \right) \cr & \Rightarrow \,\,{\cos ^{ - 1}}\left( {\frac{2}{{3x}}} \right) = {\sin ^{ - 1}}\left( {\frac{3}{{4x}}} \right) \cr & \left[ {\because \,\,{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x = \frac{\pi }{2}} \right] \cr & {\text{Put }}{\sin ^{ - 1}}\left( {\frac{3}{{4x}}} \right) = \theta \cr & \Rightarrow \,\,\sin \theta = \frac{3}{{4x}} \cr & \Rightarrow \,\,\cos \theta = \sqrt {1 - {{\sin }^2}\theta } \cr & = \sqrt {1 - \frac{9}{{16{x^2}}}} \cr & \Rightarrow \,\,\theta = {\cos ^{ - 1}}\left( {\frac{{\sqrt {16{x^2} - 9} }}{{4x}}} \right) \cr & \therefore \,\,{\cos ^{ - 1}}\left( {\frac{2}{{3x}}} \right) \cr & = {\cos ^{ - 1}}\left( {\frac{{\sqrt {16{x^2} - 9} }}{{4x}}} \right) \cr & \Rightarrow \,\,\frac{2}{{3x}} = \frac{{\sqrt {16{x^2} - 9} }}{{4x}} \cr & \Rightarrow \,\,{x^2} = \frac{{64 + 81}}{{9 \times 16}} \cr & \Rightarrow \,\,x = \pm \sqrt {\frac{{145}}{{144}}} \cr & \Rightarrow \,\,x = \frac{{\sqrt {145} }}{{12}}\,\,\left( {\because \,\,x > \frac{3}{4}} \right) \cr} $$