If `cos(sin^(-1)'2/5 + cos^(-1)x) = 0`, then x is equal to

Options

A. `1/5`
B. `2/5`
C. `0`
D. `1`

Explanation

To solve the equation \( \cos(\sin^{-1} \frac{2}{5} + \cos^{-1} x) = 0 \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \cos(\sin^{-1} \frac{2}{5} + \cos^{-1} x) = 0 \]
This implies that:
\[ \sin^{-1} \frac{2}{5} + \cos^{-1} x = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]
For simplicity, we can consider \( n = 0 \):
\[ \sin^{-1} \frac{2}{5} + \cos^{-1} x = \frac{\pi}{2} \]

### Step 2: Use the identity
We know that:
\[ \cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x \]
Substituting this into our equation gives:
\[ \sin^{-1} \frac{2}{5} + \left(\frac{\pi}{2} - \sin^{-1} x\right) = \frac{\pi}{2} \]

### Step 3: Simplify the equation
Now, we can simplify the equation:
\[ \sin^{-1} \frac{2}{5} - \sin^{-1} x = 0 \]
This leads to:
\[ \sin^{-1} \frac{2}{5} = \sin^{-1} x \]

### Step 4: Equate the arguments
Since the inverse sine function is one-to-one in the range \([-1, 1]\), we can equate the arguments:
\[ x = \frac{2}{5} \]

### Final Answer
Thus, the value of \( x \) is:
\[ \boxed{\frac{2}{5}} \]

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