given equation can be written as

$3 \,sin\, \theta – 4sin^3\theta = 4sin\theta \,sin\, 2\theta \,sin4\theta$

hence either $sin\, \theta = 0 \Rightarrow\, \theta = n\pi$

or        $3 – 4sin^2\theta = 4\, sin\, 2\theta \,sin \,4\theta$

           $3 – 2 (1 – cos\, 2\theta ) = 2 (cos\, 2\theta – cos \,6\theta )$

or        $1 = – 2 cos\, 6\theta$

            $cos \,6\theta = – \frac{1}{2} = cos\, \frac{2 \pi}{3}$

            $6\theta = 2n \pi ± \frac{2 \pi}{3}$

if $0\, \le\, \theta\, \le\, \pi$ then total solution are $0,\frac{\pi }{9},\frac{{2\pi }}{9},\frac{{4\pi }}{9},\frac{{5\pi }}{9},\frac{{7\pi }}{9},\frac{{8\pi }}{9},\pi $ is $8$ real solutions.