Question

The period of the function $\sin \left(\frac{2x}{3}\right)+\sin \left(\frac{3x}{2}\right)$ is

• A. $2\pi$
• B. $10\pi$
• C. $6\pi$
• D. $12\pi$

Answer 1

Answer: (D)

As $\sin \frac{2x}{3}=\sin \left(2\pi +\frac{2x}{3}\right)$
$=\sin \left(\frac{2}{3}\left(3\pi +x\right)\right),$ therefore , period of $\sin \frac{2x}{3}$ is $3\pi$ and also
$\sin \frac{3x}{2}=\sin \left(2\pi +\frac{3x}{2}\right)=\sin \left(\frac{3}{2}\left(\frac{4\pi }{3}+x\right)\right)$ therefore, period of $\sin \frac{3x}{2}$ is $\frac{4\pi }{3}$.
Hence penod of f(x) is L.C.M. of $3\pi$ and $\frac{4\pi }{3},$ i.e.,$12\pi$.
($\because$ set of multiples of $3\pi =3\pi ,6\pi ,9\pi ,12\pi ,.....)$ and set of multiplies of $\frac{4\pi }{3}$
$=\left\{\frac{4\pi }{3},\frac{8\pi }{3},\frac{12\pi }{3},\frac{16\pi }{3}m,...\right\}$