Find the general solution of $\theta$ for which the quadratic equation

$$\left(\sin\theta\right)x^2+(2\cos\theta)x+\dfrac{\cos\theta+\sin\theta}{2}$$ is the square of a linear function.

$$D=0$$ $$4\cos^2\theta-2\sin\theta\left(\sin\theta+\cos\theta\right)=0$$ $$2\cos^2\theta-\sin^2\theta-\sin\theta\cos\theta=0$$ $$2\cos^2\theta-2\sin\theta\cos\theta+\sin\theta\cos\theta-\sin^2\theta=0$$ $$2\cos\theta(\cos\theta-\sin\theta)+\sin\theta(\cos\theta-\sin\theta)=0$$ $$(\cos\theta-\sin\theta)(2\cos\theta+\sin\theta)=0$$ $$\tan\theta=1 \text { or } \tan\theta=-2$$

$$\theta=n\pi+\dfrac{\pi}{4} \text { or } \theta=\tan(2\pi-\tan^{-1}(2))$$ $$\theta=n\pi+\dfrac{\pi}{4} \text { or } \theta=n\pi+2\pi-\tan^{-1}(2)$$ $$\theta=n\pi+\dfrac{\pi}{4} \text { or } \theta=\pi(n+2)-\tan^{-1}(2)$$ $$\sin\theta\ne 0$$ $$\theta\ne m\pi \text { where $m \in$ I }$$

$$\theta=n\pi+\dfrac{\pi}{4} \text { can't be integral multiple of $\pi$ as } \theta=\dfrac{\pi(4n+1)}{4}$$

$$\theta=n\pi+\dfrac{\pi}{4} \text { is the valid solution }$$ $$\theta=\pi(n+2)-\tan^{-1}(2) \text { cannot be the integral multiple of $\pi$ as $\tan^{-1}(2)$ is not the integral multiple of $\pi$ } $$ $$\theta=\pi(n+2)-\tan^{-1}(2) \text { is the valid solution }$$

Hence $\theta=\pi(n+2)-\tan^{-1}(2) \text { or } \theta=n\pi+\dfrac{\pi}{4}$

But actual answer is $\theta= 2n \pi+\dfrac{\pi}{4} \text{or } \theta =(2n+1)\pi - \tan^{-1}(2) \text { where $n \in$ I}$

I tried to find out the mistake but didn't get any breakthrough. What am I missing.


2 Answers 2


Your problem is that the discriminant tells you when there is a double root, not when the polynomial is a square. When $\sin \theta$ is negative the polynomial factorises in the form $-(ax+b)^2$ and is the negative of a square.

  • $\begingroup$ This means that the valid solutions require $\sin x\ge 0$? $\endgroup$
    – user
    Nov 2, 2019 at 8:03
  • 1
    $\begingroup$ @user Indeed, because only if the leading coefficient is positive can the quadratic be a square. $\endgroup$ Nov 2, 2019 at 8:20
  • $\begingroup$ why $\sin\theta>0$, suppose it is negative and we have the following equation $-\dfrac{x^2}{\sqrt{2}}-\dfrac{2x}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}$, then we can write it as $\left(-\sqrt{2}x-\sqrt{2}\right)^2$, so we can see that it is the square of a linear function. $\endgroup$ Nov 2, 2019 at 12:59
  • $\begingroup$ @user3290550 The square of a negative number is positive $\endgroup$ Nov 2, 2019 at 13:02
  • $\begingroup$ sorry I am also trying to learn here, but how does it matter, can't we write the quadratic equation as square of a linear function if coefficient of $x^2$ is negative, in the above case we are able to write as you can see $\endgroup$ Nov 2, 2019 at 13:04

By tangent half angle formulas, by $t=\tan \theta$, we have that

$$4\cos^2\theta-2\sin\theta\left(\sin\theta+\cos\theta\right)=0 \iff 4 \cos^2 \theta-2\sin^2 \theta-2\sin \theta \cos \theta=0$$

$$6\frac{1+\cos (2\theta)}2-\sin (2\theta)-2=0$$

$$3\cos (2\theta)-\sin (2\theta)+1=0$$

$$3\frac{1-t^2}{1+t^2}- \frac{2t}{1+t^2}+1=0$$

$$3-3t^2-2t+1+t^2=0 \iff t^2+t-2=0 \iff (t-1)(t+2)=0$$

and by the original equation we also need $\sin \theta \ge 0$, therefore the solutions are

$$\theta=\frac \pi 4 +2k\pi \quad \lor \quad \theta=\pi -\arctan(2) +2k\pi$$

  • $\begingroup$ I don't get you: didn't the OP already get to $\tan \theta = 1$ or $\tan \theta = -2$? $\endgroup$
    – Toby Mak
    Nov 2, 2019 at 7:47
  • $\begingroup$ @TobyMak I've given a different way to obtain the condition which confirms the result obtained inthe first part. I'll add something on the main issue. $\endgroup$
    – user
    Nov 2, 2019 at 7:51

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