The number of real values of x satisfying `tan^-1(x/(1-x^2))+tan^-1 (1/x^3)` =(3pi)/(4)`

Explanation

To solve the equation

\[ \tan^{-1}\left(\frac{x}{1-x^2}\right) + \tan^{-1}\left(\frac{1}{x^3}\right) = \frac{3\pi}{4}, \]

we can use the formula for the sum of inverse tangents:

\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right), \]

provided that \(ab < 1\).

### Step 1: Identify \(a\) and \(b\)

Let

\[ a = \frac{x}{1-x^2} \quad \text{and} \quad b = \frac{1}{x^3}. \]

### Step 2: Apply the sum formula

Using the formula, we can rewrite the left-hand side:

\[ \tan^{-1}\left(\frac{x}{1-x^2}\right) + \tan^{-1}\left(\frac{1}{x^3}\right) = \tan^{-1}\left(\frac{\frac{x}{1-x^2} + \frac{1}{x^3}}{1 - \frac{x}{1-x^2} \cdot \frac{1}{x^3}}\right). \]

### Step 3: Simplify the numerator

The numerator becomes:

\[ \frac{x}{1-x^2} + \frac{1}{x^3} = \frac{x^4 + 1 - x^2}{x^3(1-x^2)} = \frac{x^4 - x^2 + 1}{x^3(1-x^2)}. \]

### Step 4: Simplify the denominator

The denominator simplifies to:

\[ 1 - \frac{x}{1-x^2} \cdot \frac{1}{x^3} = 1 - \frac{1}{x^2(1-x^2)} = \frac{x^2(1-x^2) - 1}{x^2(1-x^2)} = \frac{x^2 - x^4 - 1}{x^2(1-x^2)}. \]

### Step 5: Combine the results

Putting it all together, we have:

\[ \tan^{-1}\left(\frac{x^4 - x^2 + 1}{x^2 - x^4 - 1}\right) = \frac{3\pi}{4}. \]

### Step 6: Set the argument equal to \(-1\)

Since \(\tan\left(\frac{3\pi}{4}\right) = -1\), we set the argument equal to \(-1\):

\[ \frac{x^4 - x^2 + 1}{x^2 - x^4 - 1} = -1. \]

### Step 7: Cross-multiply and simplify

Cross-multiplying gives:

\[ x^4 - x^2 + 1 = - (x^2 - x^4 - 1). \]

This simplifies to:

\[ x^4 - x^2 + 1 = -x^2 + x^4 + 1. \]

### Step 8: Solve for \(x\)

This simplifies to:

\[ 0 = 0, \]

which is always true. However, we must consider the restrictions from the original equation.

### Step 9: Check for restrictions

From the original equation, we have:

1. \(1 - x^2 \neq 0 \Rightarrow x \neq \pm 1\).
2. \(x^3 \neq 0 \Rightarrow x \neq 0\).

### Conclusion

Since the equation holds for all \(x\) except \(x = 1\), \(x = -1\), and \(x = 0\), we conclude that there are no real solutions that satisfy the original equation.

### Final Answer

Thus, the number of real values of \(x\) satisfying the equation is:

**No solution.**

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