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

### Step 1: Rewrite the equation
We start with the equation:
\[ \tan^2 x - \sec^{10} x + 1 = 0 \]
We can use the identity \( \sec^2 x = 1 + \tan^2 x \) to express \( \sec^{10} x \) in terms of \( \tan^2 x \).

### Step 2: Substitute \( \sec^2 x \)
From the identity, we have:
\[ \sec^2 x = 1 + \tan^2 x \implies \sec^{10} x = (1 + \tan^2 x)^5 \]
Thus, substituting this into our equation gives:
\[ \tan^2 x - (1 + \tan^2 x)^5 + 1 = 0 \]

### Step 3: Let \( y = \tan^2 x \)
Let \( y = \tan^2 x \). The equation now becomes:
\[ y - (1 + y)^5 + 1 = 0 \]
This simplifies to:
\[ y - (1 + y)^5 + 1 = 0 \implies y + 1 - (1 + y)^5 = 0 \]

### Step 4: Analyze the function
We need to analyze the function:
\[ f(y) = y + 1 - (1 + y)^5 \]
We will find the number of solutions to \( f(y) = 0 \).

### Step 5: Find the derivative
To find the critical points, we compute the derivative:
\[ f'(y) = 1 - 5(1 + y)^4 \]
Setting \( f'(y) = 0 \) gives:
\[ 1 = 5(1 + y)^4 \implies (1 + y)^4 = \frac{1}{5} \implies 1 + y = \left(\frac{1}{5}\right)^{1/4} \]
Thus:
\[ y = \left(\frac{1}{5}\right)^{1/4} - 1 \]

### Step 6: Determine the behavior of \( f(y) \)
We can evaluate \( f(y) \) at various points to determine the number of roots. As \( y \to 0 \), \( f(0) = 2 - 1 = 1 \) (positive). As \( y \to \infty \), \( f(y) \to -\infty \) (negative).

### Step 7: Use the Intermediate Value Theorem
Since \( f(y) \) is continuous and changes from positive to negative, there is at least one root in \( (0, \infty) \).

### Step 8: Find the range of \( x \)
Since \( y = \tan^2 x \), we need to find how many solutions \( \tan^2 x = y \) has in the interval \( (0, 20) \). The function \( \tan x \) is periodic with a period of \( \pi \).

### Step 9: Count the solutions
The interval \( (0, 20) \) contains:
\[ \frac{20}{\pi} \approx 6.366 \text{ cycles of } \tan x \]
Each cycle contributes 2 solutions for \( \tan^2 x = y \) (one in each half of the cycle). Therefore, the total number of solutions is:
\[ 2 \times 6 = 12 \]

### Final Answer
Thus, the number of solutions of the equation \( \tan^2 x - \sec^{10} x + 1 = 0 \) for \( x \) in \( (0, 20) \) is **12**.