Number of Solutions to Inverse Cosine Equation

To find the total number of solutions to the equation $f(x) = \frac{10 - x}{10}$ where $f: [0, 4\pi] \to [0, \pi]$ is defined as $f(x) = \cos^{-1}(\cos x)$, we first analyze the piecewise linear representation of $f(x)$ over the interval $[0, 4\pi]$.

By definition of the principal value branch of $\cos^{-1}$, we have:

$$f(x) = \begin{cases} x, & x \in [0, \pi] \\ 2\pi - x, & x \in [\pi, 2\pi] \\ x - 2\pi, & x \in [2\pi, 3\pi] \\ 4\pi - x, & x \in [3\pi, 4\pi] \end{cases}$$

Let $g(x) = \frac{10 - x}{10} = 1 - \frac{x}{10}$. We now search for the solutions to $f(x) = g(x)$ in each sub-interval:

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Interval 1: $x \in [0, \pi]$

Here, $f(x) = x$. Solving $f(x) = g(x)$: $x = 1 - \frac{x}{10} \implies \frac{11}{10}x = 1 \implies x_1 = \frac{10}{11}$

Since $0 < \frac{10}{11} < 1 < \pi$, this root lies within $[0, \pi]$.
Thus, there is $1$ solution in this interval.

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Interval 2: $x \in [\pi, 2\pi]$

Here, $f(x) = 2\pi - x$. Solving $f(x) = g(x)$: $2\pi - x = 1 - \frac{x}{10} \implies \frac{9}{10}x = 2\pi - 1 \implies x_2 = \frac{10(2\pi - 1)}{9}$

Checking if $x_2 \in [\pi, 2\pi]$:

1. Lower bound: $\frac{10(2\pi - 1)}{9} \ge \pi \iff 20\pi - 10 \ge 9\pi \iff 11\pi \ge 10$, which holds since $\pi \approx 3.1416$.
2. Upper bound: $\frac{10(2\pi - 1)}{9} \le 2\pi \iff 20\pi - 10 \le 18\pi \iff 2\pi \le 10 \iff \pi \le 5$, which is true.

Thus, there is $1$ solution in this interval.

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Interval 3: $x \in [2\pi, 3\pi]$

Here, $f(x) = x - 2\pi$. Solving $f(x) = g(x)$: $x - 2\pi = 1 - \frac{x}{10} \implies \frac{11}{10}x = 2\pi + 1 \implies x_3 = \frac{10(2\pi + 1)}{11}$

Checking if $x_3 \in [2\pi, 3\pi]$:

1. Lower bound: $\frac{10(2\pi + 1)}{11} \ge 2\pi \iff 20\pi + 10 \ge 22\pi \iff 10 \ge 2\pi \iff \pi \le 5$, which is true.
2. Upper bound: $\frac{10(2\pi + 1)}{11} \le 3\pi \iff 20\pi + 10 \le 33\pi \iff 10 \le 13\pi$, which is clearly true.

Thus, there is $1$ solution in this interval.

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Interval 4: $x \in [3\pi, 4\pi]$

Here, $f(x) = 4\pi - x$. Solving $f(x) = g(x)$: $4\pi - x = 1 - \frac{x}{10} \implies \frac{9}{10}x = 4\pi - 1 \implies x_4 = \frac{10(4\pi - 1)}{9}$

Checking if $x_4 \in [3\pi, 4\pi]$: $x_4 \le 4\pi \iff \frac{10(4\pi - 1)}{9} \le 4\pi \iff 40\pi - 10 \le 36\pi \iff 4\pi \le 10$ However, $4\pi \approx 12.566 > 10$, so $x_4 > 4\pi$.

Furthermore, $g(x) = 1 - \frac{x}{10} \le 0$ for all $x \ge 10$. Since $3\pi \approx 9.425$, for $x \in [10, 4\pi]$, $g(x) \le 0$ whereas $f(x) > 0$. On $[3\pi, 10]$, $f(x) > 4\pi - 10 \approx 2.566$ while $g(x) < g(3\pi) \approx 0.058$, so $f(x) > g(x)$ throughout $[3\pi, 4\pi]$.

Thus, there are $0$ solutions in this interval.

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Conclusion

Summing the solutions across all four sub-intervals: $\text{Total number of solutions} = 1 + 1 + 1 + 0 = 3$