I have the following question:

Find the real values of $a$ for which the equation $$(1+\tan^2\theta)^2 + 4a\tan\theta(\tan^2\theta + 1) + 16\tan^2\theta = 0$$ has four distinct real roots in $\left(0, \dfrac{\pi}{2}\right)$.

I tried to solve the above equation by dividing the entire equation by $\tan^2\theta$ and then substituting $\tan\theta + \dfrac{1}{\tan\theta}$ as $y$ and then solving for $y$. Then I tried to apply the inequality $\tan\theta + \dfrac{1}{\tan\theta} \geqslant 2$ but couldn't find a proper range of values of $a$.

Please help. Please point out if there is any mistake in my work. Thanks in advance.


3 Answers 3


You have $y^2+4ay+16=0$, which gives $y=-2a\pm2\sqrt{a^2-4}$. Note that $\tan\theta>0$ for $\displaystyle \theta\in\left(0,\frac{\pi}{2}\right)$. The equation has four real roots in $\displaystyle \left(0,\frac{\pi}{2}\right)$ if $-2a+2\sqrt{a^2-4}>2$ and $-2a-2\sqrt{a^2-4}>2$.

As $\sqrt{a^2-4}$ is real and $a<-1-\sqrt{a^2-4}$, $a$ is negative. So, $a\le-2$.

If $a=-2$, then $-2a+2\sqrt{a^2-4}=-2a-2\sqrt{a^2-4}$ and the equation will not have $4$ real roots on the interval. So $a<-2$

$\displaystyle \begin{cases} a<-2\\ -2a-2\sqrt{a^2-4}>2\end{cases}$ $\implies$ $\displaystyle \begin{cases} a<-2\\ -a-1>\sqrt{a^2-4}\end{cases}$ $\implies$ $\displaystyle \begin{cases} a<-2\\ a^2+2a+1>a^2-4\end{cases}$ $\implies$ $\displaystyle a>\frac{-5}{2}$

We can check that when $\displaystyle -\frac{5}{2}<a<-2$, $-2a+2\sqrt{a^2-4}>2$ and $-2a-2\sqrt{a^2-4}>2$.


$$\sec^4t+4a\frac{\sin t}{\cos^3t}+\dfrac{16\sin^2t}{\cos^2t}=0$$

$$0=1+4a\sin t\cos t+16(\sin^2t\cos^2t)=1+2a\sin2t+4\sin^22t$$


As $0<2t<\pi,\sin2t>0$ $$\dfrac{1+4\sin^22t}{2\sin2t}=\dfrac1{2\sin2t}+2\sin2t\ge2\sqrt{\dfrac1{2\sin2t}\cdot2\sin2t}=2$$

$\implies -a\ge2\iff a\le-2$

But if $a=-2,\sin2t=\dfrac12,\dfrac12\implies a\ne-2$


The four roots are:





and become real when:





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