If the equation $\sqrt{2} x^2 - \sqrt{3} x +k=0$ with $k$ a constant has two solutions $\sin\theta$ and $\cos\theta$ $(0\leq\theta\leq\frac{\pi}{2})$, then $k=$……

My approach is suggested below but I am not sure how to continue.

Since $\sin\theta$ and $\cos\theta$ are two solutions of the equation,

Then we have,

$\sqrt{2} \sin^2\theta - \sqrt{3} \sin\theta +k=0$ .....Equation (1)

$\sqrt{2} \cos^2\theta - \sqrt{3} \cos\theta +k=0$ .....Equation (2)

Add (2) to (1),

$\sqrt{2} (\sin^2\theta + \cos^2\theta) - \sqrt{3} (\sin\theta + \cos\theta) +2k=0$

$\sqrt{2} - \sqrt{3} (\sin\theta + \cos\theta) +2k=0$

The answer key provided is $\frac{\sqrt{2}}{4}$. I think I am probably on the right track here but not sure how I should proceed with $\sin\theta$ and $\cos\theta$ next. Please help.


3 Answers 3


$\sin\theta+\cos\theta=\dfrac{\sqrt3}{\sqrt2}$ and $\sin\theta\cos\theta=\dfrac k{\sqrt2}$.


$\dfrac32-\sqrt2 k=1$

$k=\dfrac 1{2\sqrt2}$

  • $\begingroup$ How did you know $\sin\theta+\cos\theta=\dfrac{\sqrt3}{\sqrt2}$ and $\sin\theta\cos\theta=\dfrac k{\sqrt2}$ above? $\endgroup$ Commented Jun 16, 2019 at 13:33
  • 1
    $\begingroup$ Vieta's formulas: If $\alpha$ and $\beta$ are the two roots of the quadratics equation $ax^2+bx+c=0$, then $\alpha\beta=\frac{-b}{a}$ and $\alpha\beta=\frac ca$. $\endgroup$
    – CY Aries
    Commented Jun 16, 2019 at 13:35
  • $\begingroup$ Using $\sin\theta+\cos\theta=\frac{\sqrt3}{\sqrt 2}$ and $\sqrt2-\sqrt3(\sin\theta+\cos\theta)+2k=0$, we can conclude that $\sqrt2-\sqrt3(\frac{\sqrt3}{\sqrt 2})+2k=0$ and find the value of $k$. $\endgroup$
    – CY Aries
    Commented Jun 16, 2019 at 13:37
  • $\begingroup$ Vieta's formulas are pretty new to me but thank you, you help a lot! $\endgroup$ Commented Jun 16, 2019 at 13:46
  • 1
    $\begingroup$ If $\alpha$ and $\beta$ are the roots, then $ax^2+bx+c=a(x-\alpha)(x-\beta)=ax^2-a(\alpha+\beta)x+a\alpha\beta$. By comapring the coefficients, $b=-a(\alpha+\beta)$ and $c=a\alpha\beta$. $\endgroup$
    – CY Aries
    Commented Jun 16, 2019 at 13:48

By the Viete we obtain: $$1=\left(\sqrt{\frac{3}{2}}\right)^2-\sqrt2k.$$ Can you end it now?


I'll use $y$ in place of $\theta$ . We know that sum of roots of $ax^2+bx+c$ is $\frac{-b}{a}$ $$siny+cosy=\frac{\sqrt{3}}{\sqrt{2}}$$ also $siny+cosy=\sqrt{2}sin(y+\frac{\pi}{4})$ thus $sin(y+\frac{\pi}{4})=\frac{\sqrt{3}}{2}=sin(\frac{\pi}{3})=sin(\frac{2\pi}{3})$ thus $y=\frac{\pi}{12}$ or $\frac{5\pi}{12}$ thus $sin(\frac{\pi}{12})cos(\frac{\pi}{12})=\frac{1}{2}sin(\frac{\pi}{6})=\frac{k}{\sqrt{2}}$ thus $k=\frac{sqrt{2}}{4}$


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