Use the sum of angles formula for sin: $\displaystyle \sin(x+c) = \sin(x)\cos(c) + \cos(x)\sin(c)$.
Next, match terms. You have $\displaystyle \sin(x) + \sqrt{3}\cos(x)$. The sin(x) obviously goes with sin (x), so cos(c) must have something to do with the coefficient of sin(x), which is 1. Then cos(x) obviously goes with cos(x), so sin(c) must have something to do with the coefficient of cos(x), which is $\displaystyle \sqrt{3}$. Assume that the 1 and the $\displaystyle \sqrt{3}$ are sides of a right triangle. What would the hypotenuse be? By the Pythagorean Theorem, it would be 2. So, factor out a 2: $\displaystyle 2\left(\dfrac{1}{2}\sin(x) + \dfrac{\sqrt{3}}{2}\cos(x)\right)$. Now, what angle has $\displaystyle \cos(c) = \dfrac{1}{2}$ and $\displaystyle \sin(c) = \dfrac{\sqrt{3}}{2}$?
In general let the expression be a sin P + b cos P. We can proceed by dividing and multiplying the expression by sqrt (a^2 + b^2) So we get
a sin P + b cos P = [sqrt (a^2 + b^2)] [ {a sin P + b cos P } / sqrt (a^2 + b^2) ] = [sqrt (a^2 + b^2)] [ a/{sqrt (a^2 + b^2)} sin P + b/{sqrt (a^2 + b^2)} cos P]
Now we can put a/sqrt (a^2 + b^2) = cos A then we will have b/sqrt (a^2 + b^2) = sin A and we will get
a sin P + b cos P = [sqrt (a^2 + b^2)] [ sin P cos A + sin A cos P ]= [sqrt (a^2 + b^2)] sin ( A + P)
Alternatively we can alternatively put
a/sqrt (a^2 + b^2) = sin A then we will have b/sqrt (a^2 + b^2) = cos A and we will get
a sin P + b cos P = [sqrt (a^2 + b^2)] [ sin P sin A + cos A cos P ]= [sqrt (a^2 + b^2)] cos ( P - A)