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

Question

$\sqrt2 x^2 - \sqrt3 x + k = 0$ with solutions $\sin \theta , \cos \theta $, $\enspace0\leq\theta\leq2\pi$.

$(x-\sin \theta)(x-\cos \theta)=0$

$(\sin \theta + \cos \theta) = \sqrt3/ \sqrt2$

$(\sin \theta \cdot \cos \theta) = k/\sqrt2$

But how to find $k$?

Answer 1

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

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.

Answer 2

Spelling out @TheSimpliFire's argument, $$\frac14=\frac{(\sin\theta+\cos\theta)^2-1}{2}=\sin\theta\cos\theta=\frac{k}{\sqrt{2}}\implies k=\frac{1}{\sqrt{8}}.$$