Answer
$$x=\pm \frac{\pi}{4}+2n\pi.$$
Work Step by Step
$$ \left| \begin{array}{ccc} \cos x & 0 & \sin x \\ \sin x & 0 & \cos x\\ \sin x -\cos x&1& \sin x +\cos x \end{array}\right|=-(\cos^2 x - \sin^2 x) =\sin^2x-\cos^2x.$$
Now, solving the equation $$\sin^2x-\cos^2x=0$$
by using the property
$\cos^2x+\sin^2x=1$, we have
$$\sin^2x-\cos^2x=0\Longrightarrow \sin x=\pm \frac{1}{\sqrt{2}}.$$
Hence, the solution is given by
$$x=\pm \frac{\pi}{4}+2n\pi.$$
