Given \(\sin x = \frac{3}{4}\), we use the Pythagorean identity to find \(\cos x\):

\[
\sin^2 x + \cos^2 x = 1
\]

First, calculate \(\sin^2 x\):

\[
\sin^2 x = \left(\frac{3}{4}\right)^2 = \frac{9}{16}
\]

Substitute into the identity:

\[
\frac{9}{16} + \cos^2 x = 1
\]

Solve for \(\cos^2 x\):

\[
\cos^2 x = 1 - \frac{9}{16} = \frac{16}{16} - \frac{9}{16} = \frac{7}{16}
\]

Taking the square root:

\[
\cos x = \pm \sqrt{\frac{7}{16}} = \pm \frac{\sqrt{7}}{4}
\]

Thus, the value of \(\cos x\) is \(\frac{\sqrt{7}}{4}\). So the correct answer is:

C. \(\frac{\sqrt{7}}{4}\)