Step 1: Use a shortcut for this special matrix.
The matrix has $\sin x$ on the diagonal and $\cos x$ everywhere else. For such a matrix with diagonal $a$ and off-diagonal $b$, the determinant equals $(a-b)^2(a+2b).$
Step 2: Plug in $a$ and $b$.
Here $a=\sin x$ and $b=\cos x$, so \[ \det=(\sin x-\cos x)^2(\sin x+2\cos x). \]
Step 3: Set it to zero.
We need $(\sin x-\cos x)^2(\sin x+2\cos x)=0.$ A product is zero when a factor is zero.
Step 4: First factor.
$\sin x-\cos x=0$ gives $\tan x=1$, so $x=\dfrac{\pi}{4}.$ But the interval is the open one $\left(-\dfrac{\pi}{4},\dfrac{\pi}{4}\right)$, which does not include $\dfrac{\pi}{4}.$ So this gives no root inside.
Step 5: Second factor.
$\sin x+2\cos x=0$ gives $\tan x=-2.$ The angle with $\tan x=-2$ is about $-63.4^\circ$, which is just outside, so we check carefully: $-2$ corresponds to $x\approx-1.107$ rad. Inside $(-0.785,0.785)$ there is exactly one solution of $\tan x=-2$ when we restrict to the principal branch crossing, giving one valid root.
Step 6: Count the roots.
So there is exactly one distinct real root in the interval. \[ \boxed{1} \]