Question

The number of solutions of the equation $x + 2\tan{x} = \frac{\pi}{2}$ in the interval $(0, 2\pi)$ is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (D)

3

Answer 2

Let $h(x) = x + 2\tan{x} - \frac{\pi}{2}$. We analyze the number of roots of $h(x)$ in the interval $(0, 2\pi)$. The derivative is $h'(x) = 1 + 2\sec^2{x}$. Since $\sec^2{x} \ge 1$, $h'(x) \ge 1 + 2(1) = 3 > 0$. Thus, $h(x)$ is strictly increasing in each continuous interval of its domain.

The vertical asymptotes for $\tan{x}$ in $(0, 2\pi)$ are at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$.

1. Interval $(0, \frac{\pi}{2})$:

   • As $x \to 0^+$, $h(x) \to 0 + 2(0) - \frac{\pi}{2} = -\frac{\pi}{2}$.
   • As $x \to \frac{\pi}{2}^-$, $h(x) \to \frac{\pi}{2} + 2(+\infty) - \frac{\pi}{2} = +\infty$. Since $h(x)$ is continuous and strictly increasing from a negative to a positive value, there is exactly one solution in $(0, \frac{\pi}{2})$.

2. Interval $(\frac{\pi}{2}, \frac{3\pi}{2})$:

   • As $x \to \frac{\pi}{2}^+$, $h(x) \to \frac{\pi}{2} + 2(-\infty) - \frac{\pi}{2} = -\infty$.
   • As $x \to \frac{3\pi}{2}^-$, $h(x) \to \frac{3\pi}{2} + 2(+\infty) - \frac{\pi}{2} = +\infty$. Since $h(x)$ is continuous and strictly increasing from $-\infty$ to $+\infty$, there is exactly one solution in $(\frac{\pi}{2}, \frac{3\pi}{2})$. More specifically, we can check $h(\pi) = \pi + 2\tan(\pi) - \frac{\pi}{2} = \pi + 0 - \frac{\pi}{2} = \frac{\pi}{2} > 0$. Since $h(x) \to -\infty$ as $x \to \frac{\pi}{2}^+$ and $h(\pi) > 0$, there is a root in $(\frac{\pi}{2}, \pi)$. In $(\pi, \frac{3\pi}{2})$, $h(x)$ increases from $h(\pi) = \frac{\pi}{2}$ to $+\infty$, so there are no roots in $(\pi, \frac{3\pi}{2})$.

3. Interval $(\frac{3\pi}{2}, 2\pi)$:

   • As $x \to \frac{3\pi}{2}^+$, $h(x) \to \frac{3\pi}{2} + 2(-\infty) - \frac{\pi}{2} = -\infty$.
   • At $x=2\pi$, $h(2\pi) = 2\pi + 2\tan(2\pi) - \frac{\pi}{2} = 2\pi + 0 - \frac{\pi}{2} = \frac{3\pi}{2}$. Since $h(x)$ is continuous and strictly increasing from $-\infty$ to a positive value, there is exactly one solution in $(\frac{3\pi}{2}, 2\pi)$.

In total, there are $1 + 1 + 1 = 3$ solutions in the interval $(0, 2\pi)$.