Analyzing the Trigonometric Equation

We need to find the number of solutions for the equation $\sin^7x+\cos^7x=1$ in the interval $x \in [0, 4\pi]$.

Key Properties and Inequalities

Recall the fundamental trigonometric identity: $\sin^2x+\cos^2x=1$. For any real number $x$, we know that $-1 \le \sin x \le 1$ and $-1 \le \cos x \le 1$. Since the power is odd (7), $\sin^7x$ has the same sign as $\sin x$, and $\cos^7x$ has the same sign as $\cos x$. Consider the interval where $\sin x > 0$ and $\cos x > 0$, which is the first quadrant $(0, \pi/2)$. In this interval, $0 < \sin x < 1$ and $0 < \cos x < 1$. For $0 < a < 1$, we have $a^7 < a^2$. Therefore, for $x \in (0, \pi/2)$: $\sin^7x < \sin^2x$ $\cos^7x < \cos^2x$ Adding these inequalities, we get $\sin^7x+\cos^7x < \sin^2x+\cos^2x = 1$. Thus, there are no solutions in the open interval $(0, \pi/2)$. Similar analysis shows no solutions in the open intervals $(\pi/2, \pi)$, $(\pi, 3\pi/2)$, and $(3\pi/2, 2\pi)$ because in these intervals, either $\sin^7x < \sin^2x$ or $\cos^7x < \cos^2x$ (or both potentially negative), and the sum cannot reach 1.

Identifying Potential Solutions

The equation $\sin^7x+\cos^7x=1$ can only hold true when the terms are close to their maximum possible values, specifically when $\sin x$ or $\cos x$ are equal to $0$ or $1$. Let's check these cases:

• Case 1: $\sin x = 1$. This implies $\cos x = 0$. Substituting into the equation: $1^7 + 0^7 = 1$. This is true.
• Case 2: $\cos x = 1$. This implies $\sin x = 0$. Substituting into the equation: $0^7 + 1^7 = 1$. This is true.
• Case 3: $\sin x = 0$. This implies $\cos x = \pm 1$.
  • If $\cos x = 1$, $0^7 + 1^7 = 1$. This is true (same as Case 2).
  • If $\cos x = -1$, $0^7 + (-1)^7 = -1 \ne 1$. This is not a solution.
• Case 4: $\cos x = 0$. This implies $\sin x = \pm 1$.
  • If $\sin x = 1$, $1^7 + 0^7 = 1$. This is true (same as Case 1).
  • If $\sin x = -1$, $(-1)^7 + 0^7 = -1 \ne 1$. This is not a solution.

Therefore, the solutions occur only when $(\sin x, \cos x)$ is $(1, 0)$ or $(0, 1)$.

General Solutions

The conditions are met when:

• $\sin x = 1$ and $\cos x = 0$. This happens when $x = \frac{\pi}{2} + 2k\pi$ for any integer $k$.
• $\sin x = 0$ and $\cos x = 1$. This happens when $x = 2k\pi$ for any integer $k$.

Solutions in the Interval $[0, 4\pi]$

We need to find the values of $k$ that place $x$ within the interval $[0, 4\pi]$.

• For $x = 2k\pi$:
  • If $k=0$, $x = 0$.
  • If $k=1$, $x = 2\pi$.
  • If $k=2$, $x = 4\pi$.
• For $x = \frac{\pi}{2} + 2k\pi$:
  • If $k=0$, $x = \frac{\pi}{2}$.
  • If $k=1$, $x = \frac{\pi}{2} + 2\pi = \frac{5\pi}{2}$.
  • If $k=2$, $x = \frac{\pi}{2} + 4\pi = \frac{9\pi}{2}$ (This is greater than $4\pi$, so it's outside the interval).

The distinct solutions in the interval $[0, 4\pi]$ are $0, \frac{\pi}{2}, 2\pi, \frac{5\pi}{2}, 4\pi$.

Counting the Solutions

There are 5 distinct solutions.