**Hint:**
The period of the sine and cosine functions are well known to be $2\pi$ for both. Hence $\dfrac{2\pi}\lambda$ is a period of the linear combination, for the argument $\lambda x$.
Remains to show that it is the smallest.
* * *
Setting $t:=\lambda x$, let $T=\lambda X$ be the period.
$$A\cos(t+T)+B\sin(t+T)=A\cos(t)+B\sin(t)$$
implies, using the sum-to-product formula,
$$-2A\sin\left(t+\frac T2\right)\sin\left(\frac T2\right)+2A\cos\left(t+\frac T2\right)\sin\left(\frac T2\right)=0.$$
This expression is identically zero for the smallest nonzero value $T=2\pi$.