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

$$\sqrt2 x^2 - \sqrt3 x + k = 0$$ 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 $$k$$?

• Hint: $(\sin\theta+\cos\theta)^2=(\sin^2\theta+\cos^2\theta)+2(\sin\theta\cos\theta)\implies\cdots$ Commented Jul 28, 2019 at 8:16

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