If the system of equations  $\left(4\sin \theta \right)x-3y+z=0,x-\left(6\cos 2\theta \right)y+z=0 3x-12y+4z=0$ has  a nontrivial solution then a value of  $\theta$ is  ${\sin }^{-1}\left(\frac{a}{b}\right)$ where  $(a,b)=1$, then  $a+b=$

$$\begin{gathered} 64{\sin }^3\theta -12{\sin }^2\theta -16\sin \theta +3=0 \\ 4{\sin }^2\theta (16\sin \theta -3)-(16\sin \theta -3)=0 \\ (16\sin \theta -3)(4{\sin }^2\theta -1)=0 \\ \sin \theta =\frac{3}{16}\Rightarrow \theta ={\sin }^{-1}\left(\frac{3}{16}\right) \\ \therefore a+b=19 \end{gathered}$$$\left|\begin{array}{ccc}4\sin \theta & -3& 1\\ 1& -6\cos 2\theta & 1\\ 3& -12& 4\end{array}\right|=0$

$4\sin \theta (-24\cos 2\theta +12)+3+1(-12+18\cos 2\theta )=0$

Expanding