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 will follow these steps:

### Step 1: Understand the equation
We know that \( \cos(A + B) = 0 \) implies \( A + B = \frac{\pi}{2} + n\pi \) for some integer \( n \). In our case, we can set \( A = \sin^{-1} \frac{2}{5} \) and \( B = \cos^{-1} x \).

### Step 2: Set up the equation
From the previous step, we can write:
\[ \sin^{-1} \frac{2}{5} + \cos^{-1} x = \frac{\pi}{2} \]

### Step 3: Isolate \( \cos^{-1} x \)
Rearranging the equation gives us:
\[ \cos^{-1} x = \frac{\pi}{2} - \sin^{-1} \frac{2}{5} \]

### Step 4: Use the co-function identity
Using the identity \( \cos^{-1} x = \sin^{-1}(\sqrt{1 - x^2}) \), we can rewrite the right-hand side:
\[ \cos^{-1} x = \sin^{-1} \left( \sqrt{1 - x^2} \right) \]
Thus, we have:
\[ \sin^{-1} \left( \sqrt{1 - x^2} \right) = \frac{\pi}{2} - \sin^{-1} \frac{2}{5} \]

### Step 5: Apply the sine function
Taking the sine of both sides gives:
\[ \sqrt{1 - x^2} = \cos\left(\sin^{-1} \frac{2}{5}\right) \]

### Step 6: Calculate \( \cos(\sin^{-1} \frac{2}{5}) \)
Using the Pythagorean identity, we find:
\[ \cos(\sin^{-1} y) = \sqrt{1 - y^2} \]
Thus:
\[ \cos\left(\sin^{-1} \frac{2}{5}\right) = \sqrt{1 - \left(\frac{2}{5}\right)^2} = \sqrt{1 - \frac{4}{25}} = \sqrt{\frac{21}{25}} = \frac{\sqrt{21}}{5} \]

### Step 7: Set up the equation for \( x \)
Now we have:
\[ \sqrt{1 - x^2} = \frac{\sqrt{21}}{5} \]

### Step 8: Square both sides
Squaring both sides results in:
\[ 1 - x^2 = \frac{21}{25} \]

### Step 9: Solve for \( x^2 \)
Rearranging gives:
\[ x^2 = 1 - \frac{21}{25} = \frac{25}{25} - \frac{21}{25} = \frac{4}{25} \]

### Step 10: Find \( x \)
Taking the square root of both sides yields:
\[ x = \pm \frac{2}{5} \]
Since \( \cos^{-1} x \) is defined for \( x \) in the range [0, 1], we take:
\[ x = \frac{2}{5} \]

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