Given :
$\sin 9\theta=\sin \theta$
$\sin 9\theta-\sin \theta=0$
$2\cos \large\frac{9\theta+\theta}{2}$$\sin \large\frac{9\theta-\theta}{2}$
$\cos 5\theta-\sin 4\theta=0$
$\cos 5\theta=0$
$5\theta=\large\frac{(2n+1)\pi}{2}$
$\theta=\large\frac{(2n+1)\pi}{10}$
$\sin 4\theta=0=\sin \pi$
$4\theta=n\pi$
$\theta=\large\frac{n\pi}{4}$
$\therefore \theta=\large\frac{n\pi}{4},\frac{(2n+1)\pi}{10}$
Where $n=0,\pm 1,\pm 2....$
Hence (A) is the correct answer.