We can rewrite the given equation x + 2tan x = (π)/(2) by isolating the trigonometric function tan x. Subtracting x from both sides and dividing by 2 gives tan x = (1)/(2)((π)/(2) - x) = (π - 2x)/(4). Thus, the solutions correspond to the points of intersection between the curve y = tan x and the line y = (π - 2x)/(4) on the interval [0, 2π].

We can rewrite the given equation x + 2tan x = (π)/(2) by isolating the trigonometric function tan x. Subtracting x from both sides and dividing by 2 gives tan x = (1)/(2)((π)/(2) - x) = (π - 2x)/(4). Thus, the solutions correspond to the points of intersection between the curve y = tan x and the line y = (π - 2x)/(4) on the interval [0, 2π].

The tangent function tan x has vertical asymptotes at x = (π)/(2) and x = (3π)/(2). Therefore, the domain [0, 2π] splits naturally into three continuous intervals: [0, (π)/(2)), ((π)/(2), (3π)/(2)), and ((3π)/(2), 2π]. On each interval, y_1 = tan x is strictly increasing, while the line y_2 = (π)/(4) - (x)/(2) is strictly decreasing.

The tangent function tan x has vertical asymptotes at x = (π)/(2) and x = (3π)/(2). Therefore, the domain [0, 2π] splits naturally into three continuous intervals: [0, (π)/(2)), ((π)/(2), (3π)/(2)), and ((3π)/(2), 2π]. On each interval, y_1 = tan x is strictly increasing, while the line y_2 = (π)/(4) - (x)/(2) is strictly decreasing.

In [0, (π)/(2)), y_1 increases from 0 to ∞ while y_2 decreases from (π)/(4) ≈ 0.785 to 0, yielding exactly 1 root. In ((π)/(2), (3π)/(2)), y_1 increases continuously from -∞ to ∞, so it intersects the line exactly once. In ((3π)/(2), 2π], y_1 increases from -∞ to 0 while y_2 takes values in [-(3π)/(4), -(π)/(2)) ≈ [-2.36, -1.57]; since y_1(2π) = 0 > y_2(2π) = -(3π)/(4), the Intermediate Value Theorem ensures exactly 1 intersection here as well.

In [0, (π)/(2)), y_1 increases from 0 to ∞ while y_2 decreases from (π)/(4) ≈ 0.785 to 0, yielding exactly 1 root. In ((π)/(2), (3π)/(2)), y_1 increases continuously from -∞ to ∞, so it intersects the line exactly once. In ((3π)/(2), 2π], y_1 increases from -∞ to 0 while y_2 takes values in [-(3π)/(4), -(π)/(2)) ≈ [-2.36, -1.57]; since y_1(2π) = 0 > y_2(2π) = -(3π)/(4), the Intermediate Value Theorem ensures exactly 1 intersection here as well.

Since there is exactly 1 solution in each of the 3 disjoint sub-intervals, the equation x + 2tan x = (π)/(2) has a total of 3 solutions in the interval [0, 2π].

no of solution of the equation x plus 2 tanx = pie/2 in the interval of 0 to 2 pie

Answer: 3

Step-by-step solution

Step 1: Rearrange into two intersecting functions

We can rewrite the given equation $x + 2\tan x = \frac{\pi}{2}$ by isolating the trigonometric function $\tan x$ . Subtracting $x$ from both sides and dividing by $2$ gives $\tan x = \frac{1}{2}\left(\frac{\pi}{2} - x\right) = \frac{\pi - 2x}{4}$ . Thus, the solutions correspond to the points of intersection between the curve $y = \tan x$ and the line $y = \frac{\pi - 2x}{4}$ on the interval $[0, 2\pi]$ .

Step 2: Analyze the functions on the sub-intervals

The tangent function $\tan x$ has vertical asymptotes at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$ . Therefore, the domain $[0, 2\pi]$ splits naturally into three continuous intervals: $[0, \frac{\pi}{2})$ , $(\frac{\pi}{2}, \frac{3\pi}{2})$ , and $(\frac{3\pi}{2}, 2\pi]$ . On each interval, $y_1 = \tan x$ is strictly increasing, while the line $y_2 = \frac{\pi}{4} - \frac{x}{2}$ is strictly decreasing.

Step 3: Check intersections in each interval

In $[0, \frac{\pi}{2})$ , $y_1$ increases from $0$ to $\infty$ while $y_2$ decreases from $\frac{\pi}{4} \approx 0.785$ to $0$ , yielding exactly $1$ root. In $(\frac{\pi}{2}, \frac{3\pi}{2})$ , $y_1$ increases continuously from $-\infty$ to $\infty$ , so it intersects the line exactly once. In $(\frac{3\pi}{2}, 2\pi]$ , $y_1$ increases from $-\infty$ to $0$ while $y_2$ takes values in $[-\frac{3\pi}{4}, -\frac{\pi}{2}) \approx [-2.36, -1.57]$ ; since $y_1(2\pi) = 0 > y_2(2\pi) = -\frac{3\pi}{4}$ , the Intermediate Value Theorem ensures exactly $1$ intersection here as well.

Step 4: Sum the total number of solutions

Since there is exactly $1$ solution in each of the $3$ disjoint sub-intervals, the equation $x + 2\tan x = \frac{\pi}{2}$ has a total of $3$ solutions in the interval $[0, 2\pi]$ .

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