Solve this

$\left|\begin{array}{ccc}x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x\end{array}\right|=8$

Expanding along $R_{1}$, we get

$x\left(-x^{2}-1\right)-\sin \theta(-x \sin \theta-\cos \theta)+\cos \theta(-\sin \theta+x \cos \theta)=8$

$\Rightarrow-x^{3}-x+x \sin ^{2} \theta+\sin \theta \cos \theta-\sin \theta \cos \theta+x \cos ^{2} \theta=8$

$\Rightarrow-x^{3}-x+x\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=8$

$\Rightarrow-x^{3}-x+x=8$

$\Rightarrow x^{3}+8=0$

$\Rightarrow(x+2)\left(x^{2}-2 x+4\right)=0$

$\Rightarrow x+2=0$                  $\left[\because x^{2}-2 x+4>0 \forall x\right]$

$\Rightarrow x=-2$