Solve `{cos^(-1)x}+[tan^(-1)x]=0` for real values of `x`. Where {.} and [.] are fractional part and greatest integer functions respectively.

Explanation

To solve the equation \(\{ \cos^{-1} x \} + [ \tan^{-1} x ] = 0\) for real values of \(x\), we will break down the problem step by step.

### Step 1: Understand the Functions Involved
The equation involves the fractional part function \(\{y\}\) and the greatest integer function \([y]\):
- The fractional part \(\{y\} = y - [y]\), which lies in the range \(0 \leq \{y\} < 1\).
- The greatest integer function \([y]\) gives the largest integer less than or equal to \(y\).

### Step 2: Analyze the Equation
Given that \(\{ \cos^{-1} x \} + [ \tan^{-1} x ] = 0\), we can deduce:
- Since \(\{ \cos^{-1} x \} \geq 0\) and \([ \tan^{-1} x ]\) is an integer, the only way their sum can equal zero is if both terms are zero:
\[ \{ \cos^{-1} x \} = 0 \quad \text{and} \quad [ \tan^{-1} x ] = 0 \]

### Step 3: Solve for \(\{ \cos^{-1} x \} = 0\)
The condition \(\{ \cos^{-1} x \} = 0\) implies that \(\cos^{-1} x\) is an integer. The range of \(\cos^{-1} x\) is \([0, \pi]\), so the possible integer values are \(0, 1, 2, 3\) (but only \(0\) and \(1\) are valid since \(\cos^{-1} x\) cannot exceed \(\pi\)).
- If \(\cos^{-1} x = 0\), then \(x = 1\).
- If \(\cos^{-1} x = 1\), then \(x = \cos(1)\).

### Step 4: Solve for \([ \tan^{-1} x ] = 0\)
The condition \([ \tan^{-1} x ] = 0\) means that \(0 \leq \tan^{-1} x < 1\). This implies:
- \(x\) must be in the range \(0 \leq x < \tan(1)\).

### Step 5: Combine Results
Now we combine the results from Steps 3 and 4:
1. From \(\{ \cos^{-1} x \} = 0\), we have \(x = 1\) or \(x = \cos(1)\).
2. From \([ \tan^{-1} x ] = 0\), we require \(0 \leq x < \tan(1)\).

### Step 6: Check Validity of Solutions
- For \(x = 1\):
- \(\tan^{-1}(1) = \frac{\pi}{4}\), which satisfies \([ \tan^{-1} 1 ] = 0\).
- For \(x = \cos(1)\):
- Since \(\cos(1) < 1\), we check if \(\tan^{-1}(\cos(1)) < 1\). Since \(\cos(1)\) is positive and less than \(1\), this also satisfies \([ \tan^{-1}(\cos(1)) ] = 0\).

### Conclusion
The solutions to the equation \(\{ \cos^{-1} x \} + [ \tan^{-1} x ] = 0\) for real values of \(x\) are:
\[ x = 1 \quad \text{and} \quad x = \cos(1) \]