Show that \( \sqrt{3} \cos x+\sin x=2 \cos \left(x-30^{\circ}\right) \)

1. Express the left-hand side in the form \[ \sqrt{3} \cos x + \sin x = R \cos\left(x - \phi\right). \] 2. Use the identity \[ R \cos\left(x - \phi\right) = R \cos x \cos \phi + R \sin x \sin \phi. \] By comparing coefficients, we have: \[ R \cos \phi = \sqrt{3} \quad \text{and} \quad R \sin \phi = 1. \] 3. Solve for \( R \) by squaring and adding the equations: \[ (R \cos \phi)^2 + (R \sin \phi)^2 = (\sqrt{3})^2 + 1^2. \] This gives: \[ R^2 (\cos^2 \phi + \sin^2 \phi) = 3 + 1 = 4. \] Since \( \cos^2 \phi + \sin^2 \phi = 1 \), we find: \[ R^2 = 4 \quad \text{and thus} \quad R = 2. \] 4. Determine \( \phi \) by taking the ratio: \[ \tan \phi = \frac{R \sin \phi}{R \cos \phi} = \frac{1}{\sqrt{3}}. \] Recognizing that: \[ \tan 30^\circ = \frac{1}{\sqrt{3}}, \] we have: \[ \phi = 30^\circ. \] 5. Substitute \( R \) and \( \phi \) back into the expression: \[ \sqrt{3} \cos x + \sin x = 2 \cos\left(x - 30^\circ\right). \]