$\sqrt2 x^2 - \sqrt3 x + k = 0$ with solutions $\sin \theta , \cos \theta$

$(\sin\theta+\cos\theta)^2=1+2\cos\theta\sin\theta=1+2\frac{k}{\sqrt{2}}$.

On the other hand, $(\sin\theta+\cos\theta)^2=\frac32$, hence

$$1+\frac{2k}{\sqrt{2}}=\frac32$$

namely $$\frac{2k}{\sqrt{2}}=\frac12 \quad \Longrightarrow k=\frac{\sqrt{2}}{4}$$

Notice that this solution is acceptable. Since both $\sin\theta$ and $\cos\theta$ are bounded by one, hence their product is subject to the same bound.