Solution

For a homogeneous system to have a non-trivial solution, the determinant of the coefficient matrix must be zero.

So,

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

Let $\sin\theta=s$ and $\cos\theta=c$.

Then,

$D=\left|\begin{array}{ccc}1 & s & c\ 1 & c & s\ 1 & -s & -c\end{array}\right|$

On simplifying,

$D=2(s-c)(s+c)$

For non-trivial solution,

$D=0$

So,

$2(s-c)(s+c)=0$

Hence,

$s-c=0$ or $s+c=0$

So,

$\sin\theta=\cos\theta$ or $\sin\theta=-\cos\theta$

Case 1:

$\sin\theta=\cos\theta$

$\tan\theta=1$

In $[0,2\pi]$,

$\theta=\frac{\pi}{4},\frac{5\pi}{4}$

Case 2:

$\sin\theta=-\cos\theta$

$\tan\theta=-1$

In $[0,2\pi]$,

$\theta=\frac{3\pi}{4},\frac{7\pi}{4}$

Total number of values of $\theta$ is $4$.