Write the expression as a function of $x$, with no angle measure involved
\[
\cos \left(\frac{\pi}{6}+x\right)
\]
\[
\cos \left(\frac{\pi}{6}+x\right)=
\]
(Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.)

Step 1 ：Given the expression \(\cos \left(\frac{\pi}{6}+x\right)\)

Step 2 ：We can simplify this using the cosine sum identity, which states that \(\cos(a + b) = \cos(a)\cos(b) - \sin(a)\sin(b)\). In this case, \(a = \frac{\pi}{6}\) and \(b = x\).

Step 3 ：Substituting \(a\) and \(b\) into the cosine sum identity, we get \(\cos \left(\frac{\pi}{6}+x\right) = \cos\left(\frac{\pi}{6}\right)\cos(x) - \sin\left(\frac{\pi}{6}\right)\sin(x)\)

Step 4 ：Since \(\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\) and \(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\), we can substitute these values into the equation to get \(\cos \left(\frac{\pi}{6}+x\right) = \frac{\sqrt{3}}{2}\cos(x) - \frac{1}{2}\sin(x)\)

Step 5 ：\(\boxed{\cos \left(\frac{\pi}{6}+x\right) = \frac{\sqrt{3}}{2}\cos(x) - \frac{1}{2}\sin(x)}\)