If \(x = \frac{\pi}{6}\), find the value of \(\frac{2cosx + sinx}{sinx}\)

Step 1 ：Substitute \(x = \frac{\pi}{6}\) into the given equation, we get: \(\frac{2cos\frac{\pi}{6} + sin\frac{\pi}{6}}{sin\frac{\pi}{6}}\)

Step 2 ：Using the trigonometric values for \(\frac{\pi}{6}\), we substitute \(cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}\) and \(sin\frac{\pi}{6} = \frac{1}{2}\) into the equation, we get: \(\frac{2*\frac{\sqrt{3}}{2} + \frac{1}{2}}{\frac{1}{2}}\)

Step 3 ：Simplify the equation to get: \(\frac{\sqrt{3} + \frac{1}{2}}{\frac{1}{2}}\)

Step 4 ：Divide each term in the numerator by \(\frac{1}{2}\) to get: \(2\sqrt{3} + 1\)