The solution of equation `sin^(50)x-cos^(50)x=1` can be `(n in I)`

Options

A. `npi+(pi)/(2)`
B. `2npi +(pi)/(3)`
C. `nx+(pi)/(4)`
D. `npi +(pi)/(3)`

Answer

To solve the equation \( \sin^{50} x - \cos^{50} x = 1 \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ \sin^{50} x - \cos^{50} x = 1 \]
Rearranging gives us:
\[ \sin^{50} x = 1 + \cos^{50} x \]

### Step 2: Analyze the ranges of sine and cosine
We know that the range of \( \sin x \) and \( \cos x \) is between -1 and 1. Therefore, \( \sin^{50} x \) will also be in the range [0, 1] since raising to an even power (like 50) will yield non-negative results.

### Step 3: Set up conditions
Since \( \sin^{50} x \) must equal \( 1 + \cos^{50} x \), we can see that:
- The maximum value of \( \sin^{50} x \) is 1 (when \( \sin x = 1 \)).
- The minimum value of \( \cos^{50} x \) is 0 (when \( \cos x = 0 \)).

This implies that:
\[ 1 + \cos^{50} x \geq 1 \]
Thus, \( \sin^{50} x \) can only equal 1 when \( \cos^{50} x = 0 \).

### Step 4: Solve for \( \cos x = 0 \)
The equation \( \cos^{50} x = 0 \) leads us to:
\[ \cos x = 0 \]
The cosine function equals zero at:
\[ x = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]
This means the solutions are at odd multiples of \( \frac{\pi}{2} \).

### Step 5: Final solution
Thus, the complete solution for the equation \( \sin^{50} x - \cos^{50} x = 1 \) is:
\[ x = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]