Given equation, \[\mathbf{cos}\text{ }(\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}~}}\mathbf{x})\text{ }=\text{ }\mathbf{sin}\text{ }(\mathbf{co}{{\mathbf{t}}^{-\mathbf{1}~}}\mathbf{3}/\mathbf{4})\]
Taking L.H.S,
On squaring on both sides,
\[\begin{array}{*{35}{l}}
16({{x}^{2}}~+\text{ }1)\text{ }=\text{ }25 \\
16{{x}^{2}}~+\text{ }16\text{ }=\text{ }25 \\
16{{x}^{2}}~=\text{ }9 \\
{{x}^{2}}~=\text{ }9/16 \\
\end{array}\]
Therefore, \[x\text{ }=\text{ }\pm \text{ }3/4\]