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

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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:
   $$
   \sqrt{3} \cos x + \sin x = 2 \cos\left(x - 30^\circ\right).
   $$
$$ \sqrt{3} \cos x + \sin x = 2 \cos\left(x - 30^\circ\right) $$

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

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