First, you find the values of $\sin\theta$ that solve the equation, using the quadratic formula or some other method. Then you use inverse trigonometric functions to find the values of $\theta$.
Setting $x=\sin(\theta)$, we have to solve $x^2 -x -1 =0.$ Since $x^2-x-1 = \left(x - \frac{1+\sqrt{5}}{2}\right)\left(x - \frac{1-\sqrt{5}}{2}\right),$ then the two solutions are $x = \frac{1+\sqrt{5}}{2},\quad\text{and}\quad \frac{1-\sqrt{5}}{2}.$
Now, $1+\sqrt{5} \gt 1+2 = 3$, so $\sin(\theta) = \frac{1+\sqrt{5}}{2}$ has no solutions. Thus, the only possibility is for $\sin(\theta) = \frac{1-\sqrt{5}}{2}$.
That means that the principal value of $\theta$ that solves the equation is $\theta = \arcsin\left(\frac{1-\sqrt{5}}{2}\right).$ And of course, you also need to consider other values of $\theta$ outside of $[-\pi/2,\pi/2]$ that may yield the same value of $\theta$. There should be at least one other on $[-\pi,\pi]$, and then you can also add multiples of $2\pi$.