Analyzing the Equation
The problem requires finding the general solution for the trigonometric equation:
$ \sin^{100} x - \cos^{100} x = 1 $We know the bounds for sine and cosine functions:
$ -1 \le \sin x \le 1 \quad \text{and} \quad -1 \le \cos x \le 1 $Since the powers are even (100), the terms $ \sin^{100} x $ and $ \cos^{100} x $ are non-negative:
$ 0 \le \sin^{100} x \le 1 \quad \text{and} \quad 0 \le \cos^{100} x \le 1 $For the equation $ \sin^{100} x - \cos^{100} x = 1 $ to hold true, given these constraints, the following conditions must be met simultaneously:
- $ \sin^{100} x = 1 $
- $ \cos^{100} x = 0 $
The first condition $ \sin^{100} x = 1 $ implies $ \sin x = \pm 1 $. The second condition $ \cos^{100} x = 0 $ implies $ \cos x = 0 $.
Deriving the General Solution
We need to find angles $x$ where both $ \cos x = 0 $ and $ \sin x = \pm 1 $ are satisfied.
The equation $ \cos x = 0 $ is satisfied when $x$ is an odd multiple of $ \frac{\pi}{2} $. The general form for these angles is:
$ x = n\pi + \frac{\pi}{2}, \quad \text{where } n \in I $(Here, $I$ denotes the set of integers).
Let's verify the value of $ \sin x $ for these angles:
- If $n$ is even ($n=2k$ for some integer $k$), then $x = 2k\pi + \frac{\pi}{2}$. In this case, $ \sin x = \sin(2k\pi + \frac{\pi}{2}) = \sin(\frac{\pi}{2}) = 1 $.
- If $n$ is odd ($n=2k+1$ for some integer $k$), then $x = (2k+1)\pi + \frac{\pi}{2} = 2k\pi + \frac{3\pi}{2}$. In this case, $ \sin x = \sin(2k\pi + \frac{3\pi}{2}) = \sin(\frac{3\pi}{2}) = -1 $.
In both scenarios ($ \sin x = 1 $ or $ \sin x = -1 $), $ \sin^{100} x = (\pm 1)^{100} = 1 $. Thus, the angles satisfying $ \cos x = 0 $ automatically satisfy $ \sin^{100} x = 1 $.
Therefore, the general solution is $ x = n\pi + \frac{\pi}{2} $. This set of solutions is equivalent to $ n\pi \pm \frac{\pi}{2} $.
The solution set is $ \left\{ n\pi \pm \frac{\pi}{2} : n \in I \right\} $.
Final Answer Verification
Comparing our derived solution $ \left\{ n\pi \pm \frac{\pi}{2} : n \in I \right\} $ with the given options:
- Option 1: $ \left\{ 2n\pi + \frac{\pi}{3} : n \in I \right\} $
- Option 2: $ \left\{ n\pi + \frac{\pi}{4} : n \in I \right\} $
- Option 3: $ \left\{ n\pi \pm \frac{\pi}{2} : n \in I \right\} $
- Option 4: $ \left\{ 2n\pi - \frac{\pi}{3} : n \in I \right\} $
Option 3 correctly represents the general solution set derived from the equation.
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