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)

$12 \, \pi$

Answer 2

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\\}