`sin(1/5 cos^(-1)x)=1`

Answer

To solve the equation \( \sin\left(\frac{1}{5} \cos^{-1}(x)\right) = 1 \), we can follow these steps:

### Step 1: Understanding the Sine Function
The sine function equals 1 at specific angles. The general solution for \( \sin(\theta) = 1 \) is given by:
\[ \theta = \frac{\pi}{2} + 2n\pi, \quad n \in \mathbb{Z} \]
This means that \( \theta \) can take values like \( \frac{\pi}{2}, \frac{5\pi}{2}, \frac{9\pi}{2}, \ldots \)

### Step 2: Setting Up the Equation
We set \( \theta = \frac{1}{5} \cos^{-1}(x) \). Therefore, we can write:
\[ \frac{1}{5} \cos^{-1}(x) = \frac{\pi}{2} + 2n\pi \]

### Step 3: Isolating \( \cos^{-1}(x) \)
To isolate \( \cos^{-1}(x) \), we multiply both sides of the equation by 5:
\[ \cos^{-1}(x) = 5\left(\frac{\pi}{2} + 2n\pi\right) = \frac{5\pi}{2} + 10n\pi \]

### Step 4: Analyzing the Range of \( \cos^{-1}(x) \)
The range of \( \cos^{-1}(x) \) is from \( 0 \) to \( \pi \). Therefore, we need to check if the expression \( \frac{5\pi}{2} + 10n\pi \) can fall within this range.

### Step 5: Checking for Valid \( n \)
1. For \( n = 0 \):
\[ \cos^{-1}(x) = \frac{5\pi}{2} \quad (\text{which is } \frac{5\pi}{2} > \pi) \]
2. For \( n = 1 \):
\[ \cos^{-1}(x) = \frac{5\pi}{2} + 10\pi = \frac{25\pi}{2} \quad (\text{which is also } > \pi) \]
3. For \( n = -1 \):
\[ \cos^{-1}(x) = \frac{5\pi}{2} - 10\pi = -\frac{15\pi}{2} \quad (\text{which is } < 0) \]

### Step 6: Conclusion
Since there are no integer values of \( n \) that yield a valid output for \( \cos^{-1}(x) \) within its defined range of \( [0, \pi] \), we conclude that the equation \( \sin\left(\frac{1}{5} \cos^{-1}(x)\right) = 1 \) has no solution.