The number of real solutions of the equation | x| = 1 - (x)/(4π) in...

The given equation is $|\sin x| = 1 - \dfrac{x}{4\pi}$.

Let $f(x) = |\sin x|$ and $g(x) = 1 - \dfrac{x}{4\pi}$. We need to find the number of points of intersection of these two graphs in the interval $[0, 4\pi]$.

The graph of $f(x) = |\sin x|$ consists of $4$ positive arches of a sine wave in the interval $[0, 4\pi]$, each of width $\pi$ and maximum height $1$.

The graph of $g(x) = 1 - \dfrac{x}{4\pi}$ is a straight line passing through $(0, 1)$ and $(4\pi, 0)$.

Let us analyze the intersections in each interval of length $\pi$:

1. In $[0, \pi]$: $g(x)$ decreases from $1$ to $0.75$. The maximum of $f(x)$ is $1$ at $x = \dfrac{\pi}{2}$. Since $1 > g\left(\dfrac{\pi}{2}\right) = \dfrac{7}{8}$, the line intersects the curve twice.

2. In $[\pi, 2\pi]$: $g(x)$ decreases from $0.75$ to $0.5$. The maximum of $f(x)$ is $1$ at $x = \dfrac{3\pi}{2}$. Since $1 > g\left(\dfrac{3\pi}{2}\right) = \dfrac{5}{8}$, the line intersects the curve twice.

3. In $[2\pi, 3\pi]$: $g(x)$ decreases from $0.5$ to $0.25$. The maximum of $f(x)$ is $1$ at $x = \dfrac{5\pi}{2}$. Since $1 > g\left(\dfrac{5\pi}{2}\right) = \dfrac{3}{8}$, the line intersects the curve twice.

4. In $[3\pi, 4\pi]$: $g(x)$ decreases from $0.25$ to $0$. The maximum of $f(x)$ is $1$ at $x = \dfrac{7\pi}{2}$. Since $1 > g\left(\dfrac{7\pi}{2}\right) = \dfrac{1}{8}$, the curve goes above the line. At $x = 4\pi$, $f(4\pi) = 0$ and $g(4\pi) = 0$, giving one intersection at the boundary. Since $f(x) > g(x)$ at $x = \dfrac{7\pi}{2}$ and $f(x)$ is concave down, there is exactly one intersection in $(3\pi, \dfrac{7\pi}{2})$ and one at $x = 4\pi$. Thus, there are $2$ intersections in this interval.

Total number of solutions = $2 + 2 + 2 + 2 = 8$.

Answer: $8$