Question

For $\theta \in (0,2\pi) - \{\pi\}$, sum of all values of '$\theta$' satisfying equation $4\cot^4\theta + 9\cot^3\theta + 8\cot^2\theta + 9\cot\theta + 4 = 0$; is equal to :

• A. $5\pi$
• B. $6\pi$
• C. $7\pi$
• D. None

Answer 1

Answer: (A)

$5\pi$

Answer 2

The given equation is $4\cot^4\theta + 9\cot^3\theta + 8\cot^2\theta + 9\cot\theta + 4 = 0$. The domain for $\theta$ is $(0, 2\pi) - \{\pi\}$. Let $x = \cot\theta$. The equation becomes $4x^4 + 9x^3 + 8x^2 + 9x + 4 = 0$. This is a reciprocal equation. Since $x=0$ is not a root, we can divide by $x^2$:

$4x^2 + 9x + 8 + \frac{9}{x} + \frac{4}{x^2} = 0$

$4(x^2 + \frac{1}{x^2}) + 9(x + \frac{1}{x}) + 8 = 0$

Let $y = x + \frac{1}{x}$. Then $y^2 = x^2 + 2 + \frac{1}{x^2}$, so $x^2 + \frac{1}{x^2} = y^2 - 2$. Substituting this into the equation gives:

$4(y^2 - 2) + 9y + 8 = 0$

$4y^2 - 8 + 9y + 8 = 0$

$4y^2 + 9y = 0$

$y(4y + 9) = 0$

This gives $y=0$ or $y = -\frac{9}{4}$.

Case 1: $y = 0$

$x + \frac{1}{x} = 0 \implies x^2 + 1 = 0 \implies x^2 = -1$. $x = \pm i$. Since $x = \cot\theta$, and $\cot\theta$ must be real for real $\theta$, there are no real solutions for $x$ in this case.

Case 2: $y = -\frac{9}{4}$

$x + \frac{1}{x} = -\frac{9}{4}$

$4x^2 + 4 = -9x$

$4x^2 + 9x + 4 = 0$

This is a quadratic equation for $x = \cot\theta$. The discriminant is $\Delta = 9^2 - 4(4)(4) = 81 - 64 = 17$. Since $\Delta > 0$, there are two distinct real roots for $x$. The roots are $x_1 = \frac{-9 + \sqrt{17}}{8}$ and $x_2 = \frac{-9 - \sqrt{17}}{8}$. Both roots are negative.

We need to find the values of $\theta \in (0, 2\pi) - \{\pi\}$ such that $\cot\theta = x_1$ or $\cot\theta = x_2$. Since $x_1 < 0$ and $x_2 < 0$, the values of $\theta$ must be in the second or fourth quadrant. So, the solutions for $\cot\theta = x_1$ in the given domain are $\theta_1$ and $\theta_2 = \pi + \theta_1$. Similarly, the solutions for $\cot\theta = x_2$ in the given domain are $\theta_3$ and $\theta_4 = \pi + \theta_3$.

The sum of the solutions is $S = \theta_1 + (\pi + \theta_1) + \theta_3 + (\pi + \theta_3) = 2\pi + 2\theta_1 + 2\theta_3 = 2\pi + 2(\theta_1 + \theta_3)$.

$\cot(\theta_1 + \theta_3) = \frac{\cot\theta_1 \cot\theta_3 - 1}{\cot\theta_1 + \cot\theta_3} = \frac{x_1 x_2 - 1}{x_1 + x_2}$. From the quadratic equation $4x^2 + 9x + 4 = 0$, the sum of roots $x_1+x_2 = -9/4$ and the product of roots $x_1x_2 = 4/4 = 1$.

$\cot(\theta_1 + \theta_3) = \frac{1 - 1}{-9/4} = \frac{0}{-9/4} = 0$.

Since $\theta_1, \theta_3 \in (\pi/2, \pi)$, we have $\pi/2 + \pi/2 < \theta_1 + \theta_3 < \pi + \pi$, so $\pi < \theta_1 + \theta_3 < 2\pi$. The general solution for $\cot\phi = 0$ is $\phi = n\pi + \pi/2$, $n \in \mathbb{Z}$. We need $\pi < n\pi + \pi/2 < 2\pi$, which leads to $\theta_1 + \theta_3 = \frac{3\pi}{2}$.

The sum of the solutions is $S = 2\pi + 2(\frac{3\pi}{2}) = 2\pi + 3\pi = 5\pi$.