What is the largest value of $k$ such that the equation $$x^2+4(\sin^{2}{x}\tan^{2}{x}+\cos^{2}{x}\cot^{2}{x}+k^2-x\sec{x}\csc{x})=0$$ has real roots?
I tried to find out the roots in terms of $k$. But I couldn't.
How do we approach this problem without finding the roots actually?
On
I would suggest plugging your equation into the quadratic formula to get
$$ x = \frac{1}{2} \sqrt{(-16)(\sin^2 x \tan^2 x + \cos^2 x \cot^2 x +k^2 - x \sec x \csc x)}$$
From here we see we need
$$(-16)(\sin^2 x \tan^2 x + \cos^2 x \cot^2 x +k^2 - x \sec x \csc x) \gt 0$$
to have real roots. Rearranging $\to$
$$ k^2 \lt - \sin^2 x \tan^2 x - \cos^2 x \cot^2 x + x \sec x \csc x$$
On
\begin{align} C &\equiv \cos x \\ S &\equiv \sin x \\ 0 &= x^2-(4\sec x \csc x)\, x+4(\sin^2 x \tan^2 x+\cos^2 x \cot^2 x+k^2) \\ &= x^2-\frac{4x}{SC}+4\left( \frac{S^6+C^6}{S^2 C^2}+k^2 \right) \\ \Delta &= \left( \frac{4}{SC} \right)^2- 16\left( \frac{S^6+C^6}{S^2 C^2}+k^2 \right) \\ &= 16\left[ \frac{1-(S^6+C^6)}{S^2C^2}-k^2 \right] \\ &= 16 \left[ \frac{1-(\color{red}{S^2+C^2})(S^4-S^2C^2+C^4)}{S^2C^2}-k^2 \right] \\ &= 16\left[ \frac{1-(S^4-S^2C^2+C^4)}{S^2C^2}-k^2 \right] \\ &= 16\left[ \frac{1+3S^2C^2-(\color{red}{S^2+C^2})^2}{S^2C^2}-k^2 \right] \\ &= 16(3-k^2) \\ \end{align}
Therefore,
$$\fbox{$-\sqrt{3} \le k \le \sqrt{3} \,$}$$
On
$$x^2+4(\sin^{2}{x}\tan^{2}{x}+\cos^{2}{x}\cot^{2}{x}+k^2-x\sec{x}\csc{x})=0$$
$$x^2+4((1-\cos^{2}{x})\tan^{2}{x}+(1-\sin^{2}{x})\cot^{2}{x}+k^2-\frac{x}{\sin{x}\cos{x}})=0$$
$$x^2+4(\tan^{2}{x}-\sin^{2}{x}+\cot^{2}{x}-\cos^{2}{x}+k^2-\frac{x}{\sin{x}\cos{x}})=0$$
$$x^2+4(\tan^{2}{x}+\cot^{2}{x}-1+k^2-\frac{x}{\sin{x}\cos{x}})=0$$
$$\frac{x^2}{4}+k^2-1+\tan^{2}{x}+\cot^{2}{x}-\frac{x}{\sin{x}\cos{x}}=0$$
$$\frac{x^2}{4}+k^2-1+\frac{\sin^{2}{x}}{\cos^{2}{x}}+\frac{\cos^{2}{x}}{\sin^{2}{x}}-\frac{x}{\sin{x}\cos{x}}=0$$
$$\frac{x^2}{4}+k^2-1+\frac{\sin^{4}{x}+\cos^{4}{x}}{\sin^{2}{x}\cos^{2}{x}}-\frac{x}{\sin{x}\cos{x}}=0$$
$$\frac{x^2}{4}+k^2-1+\frac{(\sin^{2}{x}+\cos^{2}{x})^2-2\sin^{2}{x}\cos^{2}{x}}{\sin^{2}{x}\cos^{2}{x}}-\frac{x}{\sin{x}\cos{x}}=0$$
$$\frac{x^2}{4}+k^2-1+\frac{1-2\sin^{2}{x}\cos^{2}{x}}{\sin^{2}{x}\cos^{2}{x}}-\frac{x}{\sin{x}\cos{x}}=0$$
$$\frac{x^2}{4}+k^2-3+\frac{1}{\sin^{2}{x}\cos^{2}{x}}-\frac{x}{\sin{x}\cos{x}}=0$$
$$\frac{x^2}{4}+k^2-3+\frac{4}{\sin^{2}{2x}}-\frac{2x}{\sin{2x}}=0$$
$$\left(\frac{2}{\sin{2x}}-\frac{x}{2}\right)^2=3-k^2$$
$$\frac{2}{\sin{2x}}-\frac{x}{2}=\pm\sqrt{3-k^2}$$
So we know $k \le \sqrt{3}$, otherwise the right hand side will not be real. When $k=\sqrt{3}$, the equation becomes $\sin{2x}=\frac{4}{x}$. This must have real solutions because $\lim_{x\to\infty}\frac{4}{x}=0$ and we know sin oscillates between -1 and 1.
$$x^2- (4\sec{x}\csc{x})x +4(\sin^{2}{x}\tan^{2}{x}+\cos^{2}{x}\cot^{2}{x}+k^2)=0$$ Look for the discriminant, for real roots $D\ge 0 $ $$16\sec^2 x\csc^2 x-16(\sin^2 x\tan^2 x+\cos^2 x\cot^2 x+k^2)\ge 0$$ $$\frac{1}{\sin^2 x\cos^2 x}-\frac{\sin^4 x}{\cos^2 x}-\frac{\cos^4 x}{\sin^2 x}-k^2\ge 0$$ $$\frac{1-\sin^6 x-\cos^6 x}{\sin^2 x\cos^2 x}\ge k^2$$ $$\frac{\sin^2 x+\cos^2 x-\sin^6 x-\cos^6 x}{\sin^2 x\cos^2 x}\ge k^2 $$ $$\frac{\sin^2 x(1-\sin^4 x)+\cos^2 x(1-\cos^4 x)}{\sin^2 x\cos^2 x}\ge k^2$$ $$\frac{\sin^2 x(1-\sin^2 x)(1+\sin^2 x)+\cos^2 x(1-\cos^2 x)(1+\cos^2 x)}{\sin^2 x\cos^2 x}\ge k^2$$ $$\frac{\sin^2 x\cos^2 x(1+\sin^2 x +1+\cos^2 x)}{\sin^2 x\cos^2 x}\ge k^2$$ $3\ge k^2$ $\implies -\sqrt3\le k\le \sqrt3\;$ So maximum value of $k$ is $\sqrt3$