If ${\cos ^{ - 1}}\left( {{2 \over {3x}}} \right) + {\cos ^{ - 1}}\left( {{3 \over {4x}}} \right) = {\pi \over 2}$ (x > $3 \over 4$), then x is equal to :

Question

${{\sqrt {145} } \over {10}}$

${{\sqrt {145} } \over {11}}$

${{\sqrt {145} } \over {12}}$

${{\sqrt {146} } \over {12}}$

Solution

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