Analyze Equation $2x + 3\tan x = \pi$ Solutions

The problem requires finding the number of solutions for the equation $2x + 3\tan x = \pi$ within the specified interval $x \in [-2\pi, 2\pi]$. The points $x = \pm \frac{\pi}{2}, \pm \frac{3\pi}{2}$ are excluded because the tangent function is undefined there.

Let's define a function $f(x) = 2x + 3\tan x - \pi$. The solutions to the original equation are the roots of $f(x) = 0$.

Derivative Analysis

To understand the behavior of $f(x)$, we compute its derivative: $ f'(x) = \frac{d}{dx}(2x + 3\tan x - \pi) $ $ f'(x) = 2 + 3\sec^2 x $

Since $\sec^2 x \ge 1$ for all $x$ where it is defined, $f'(x) \ge 2 + 3(1) = 5$. Because $f'(x)$ is always positive, the function $f(x)$ is strictly increasing on each interval where it is continuous. A strictly increasing function can cross the x-axis at most once within any continuous interval.

Interval Solution Count

We need to check the intervals defined by the excluded points within $[-2\pi, 2\pi]$: $[-2\pi, -\frac{3\pi}{2})$, $(-\frac{3\pi}{2}, -\frac{\pi}{2})$, $(-\frac{\pi}{2}, \frac{\pi}{2})$, $(\frac{\pi}{2}, \frac{3\pi}{2})$, and $(\frac{3\pi}{2}, 2\pi]$.

• Interval 1: $[-2\pi, -\frac{3\pi}{2})$

  Evaluate $f(x)$ at the endpoints/limits: $f(-2\pi) = 2(-2\pi) + 3\tan(-2\pi) - \pi = -4\pi - \pi = -5\pi$. As $x$ approaches $-\frac{3\pi}{2}$ from the left ($x \to -\frac{3\pi}{2}^-$), $\tan x \to +\infty$, thus $f(x) \to +\infty$. Since $f(x)$ increases from $-5\pi$ to $+\infty$, it crosses the x-axis once. 1 solution.

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

  Evaluate limits: As $x \to -\frac{3\pi}{2}^+$, $\tan x \to -\infty$, so $f(x) \to -\infty$. As $x \to -\frac{\pi}{2}^-$, $\tan x \to +\infty$, so $f(x) \to +\infty$. Since $f(x)$ increases from $-\infty$ to $+\infty$, it crosses the x-axis once. 1 solution.

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

  Evaluate limits: As $x \to -\frac{\pi}{2}^+$, $\tan x \to -\infty$, so $f(x) \to -\infty$. As $x \to \frac{\pi}{2}^-$, $\tan x \to +\infty$, so $f(x) \to +\infty$. Since $f(x)$ increases from $-\infty$ to $+\infty$, it crosses the x-axis once. 1 solution.

• Interval 4: $(\frac{\pi}{2}, \frac{3\pi}{2})$

  Evaluate limits: As $x \to \frac{\pi}{2}^+$, $\tan x \to -\infty$, so $f(x) \to -\infty$. As $x \to \frac{3\pi}{2}^-$, $\tan x \to +\infty$, so $f(x) \to +\infty$. Since $f(x)$ increases from $-\infty$ to $+\infty$, it crosses the x-axis once. 1 solution.

• Interval 5: $(\frac{3\pi}{2}, 2\pi]$

  Evaluate limits/endpoints: As $x \to \frac{3\pi}{2}^+$, $\tan x \to -\infty$, so $f(x) \to -\infty$. At $x = 2\pi$, $f(2\pi) = 2(2\pi) + 3\tan(2\pi) - \pi = 4\pi - \pi = 3\pi$. Since $f(x)$ increases from $-\infty$ to $3\pi$, it crosses the x-axis once. 1 solution.

Total Solutions

Adding the number of solutions from each interval: $1 + 1 + 1 + 1 + 1 = 5$.

There are a total of 5 solutions for the equation $2x + 3\tan x = \pi$ in the interval $[-2\pi, 2\pi]$ excluding the specified points.