Finding the period of $f(x) = \sin 2x + \cos 3x$ I want to find the period of the function $f(x) = \sin 2x + \cos 3x$. I tried to rewrite it using the double angle formula and addition formula for cosine. However, I did not obtain an easy function. Another idea I had was to calculate the zeros and find the difference between the zeros. But that is only applicable if the function oscillates around $y = 0$, right? My third approach was to calculate the extrema using calculus and from that derive the period. Does anybody have another approach?
Thanks
 A: In general, if $T$ is the period of a function $f(x)$ then the period of the function $f(ax)$ is $\frac{T}{a}$
In general, if two periodic functions $f_1(x)$ & $f_2(x)$ have periods $T_1$ & $T_2$ then the period of the function $g(x)=f_1(x)\pm f_2(x)$ is L.C.M. (Least common multiple) of $T_1$ & $T_2$.
As per your question, the periods of $\sin2x$ & $\cos3x$ are calculated as $\frac{2\pi}{2}=\pi$ & $\frac{2\pi}{3}$ respectively. 
Hence the period of the function $f(x)=\sin2x+\cos3x$ is L.C.M. of $\pi$ & $\frac{2\pi}{3}$ which is given by generalized formula of L.C.M. of fractions $$=\frac{\text{L.C.M. (least common multiple) of numerators}}{\text{H.C.F. (highest common factor) of denominators}}=\frac{2\pi}{1}=2\pi$$ Hence, the required period is $2\pi$
A: $\sin(2x)=\sin(2(x+k\pi))$ and
$\cos(3x)=\cos(3(x+l2\pi/3))$, then when $k=\dfrac{2l}3$, $3k=2l=6m$, the function repeats itself, and the period is at most $2\pi$.
Anyway, remains to prove that there is no shorter period.
A: Hint
The period of $\sin(2x)$ is $\pi$, and the period of $\cos(3x)$ is $2\pi/3$.
Can you find a point where both will be at the start of a new period?
