The number of solutions of the equation `tan^(2)x-sec^(10)x+1=0" for " x in (0, 20)` is equal to

Answer

To find the number of solutions for the equation \( \tan^2 x - \sec^{10} x + 1 = 0 \) in the interval \( x \in (0, 20) \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \tan^2 x - \sec^{10} x + 1 = 0 \]
Recall the identity \( \sec^2 x = 1 + \tan^2 x \). Therefore, we can express \( \sec^{10} x \) in terms of \( \tan^2 x \):
\[ \sec^{10} x = (1 + \tan^2 x)^5 \]
Substituting this into the equation gives us:
\[ \tan^2 x - (1 + \tan^2 x)^5 + 1 = 0 \]

### Step 2: Analyze the behavior of the function
To find the number of solutions, we need to analyze the behavior of the function:
\[ f(x) = \tan^2 x - (1 + \tan^2 x)^5 + 1 \]
We need to find when \( f(x) = 0 \).

### Step 3: Determine the limits of \( \sec^2 x \)
Since \( \sec^2 x \geq 1 \), we know that \( \tan^2 x \) can take values from \( 0 \) to \( \infty \). The minimum value of \( \sec^{10} x \) occurs when \( \sec^2 x = 1 \), which corresponds to \( x = n\pi \) for integers \( n \).

### Step 4: Find the critical points
The equation \( \sec^2 x = 1 \) implies \( \cos x = 1 \), which gives solutions:
\[ x = n\pi \quad \text{for integers } n \]
Now we need to find how many such solutions exist in the interval \( (0, 20) \).

### Step 5: Calculate the integer values of \( n \)
To find the values of \( n \):
- The smallest \( n \) such that \( n\pi < 20 \) can be calculated as:
\[ n < \frac{20}{\pi} \approx \frac{20}{3.14} \approx 6.37 \]
Thus, \( n \) can take values \( 1, 2, 3, 4, 5, 6 \).

### Step 6: Count the solutions
The integer values of \( n \) that satisfy this condition are \( n = 1, 2, 3, 4, 5, 6 \), which gives us a total of \( 6 \) solutions.

### Conclusion
The number of solutions of the equation \( \tan^2 x - \sec^{10} x + 1 = 0 \) for \( x \in (0, 20) \) is \( 6 \).