Width at half the maximum height of a sin-function is one-third of wavelength?

Question

For the sin finciton $f(x)=A\sin (k x)$, the maximum height is $A$ and wavelength is $\lambda=\frac{2\pi}{k}$. Is the following equation correct?

If $x_1=\arcsin(\frac{A}{2})$ for $0<x_1<\frac{\pi}{2k}$, then $2(\frac{\pi}{2k}-x_1)=\frac{\lambda}{3}$.

Answer 1

HINT:

From

$$ A \sin \frac{2 \pi x}{\lambda} =\frac{A}{2}$$

there are two values of $x_1, x_2$ to be found