Let `I=int(dx)/((cosx-sinx^(2)))=(1)/(f(x))+C` (where, C is the constant of integration). If `f((pi)/(3))=1-sqrt3`, then the number of solution(s) of `f(x)=2020` in `x in ((pi)/(2), pi)` is/are

Answer

To solve the problem step by step, we need to find the function \( f(x) \) from the given integral and then determine the number of solutions for \( f(x) = 2020 \) in the interval \( \left( \frac{\pi}{2}, \pi \right) \).

### Step 1: Set up the integral
We start with the integral:
\[ I = \int \frac{dx}{\cos x - \sin^2 x} \]
We need to simplify the integrand.

### Step 2: Simplify the integrand
Notice that we can rewrite the denominator:
\[ \cos x - \sin^2 x = \cos x - (1 - \cos^2 x) = \cos^3 x - 1 \]
Thus, we have:
\[ I = \int \frac{dx}{\cos^3 x - 1} \]

### Step 3: Use a substitution
Let \( t = \tan x \). Then, we have:
\[ dx = \frac{dt}{1 + t^2} \]
And since \( \cos x = \frac{1}{\sqrt{1+t^2}} \), we can rewrite the integral in terms of \( t \).

### Step 4: Rewrite the integral
Substituting \( \tan x = t \):
\[ \cos x = \frac{1}{\sqrt{1+t^2}}, \quad \sin x = \frac{t}{\sqrt{1+t^2}} \]
The integral becomes:
\[ I = \int \frac{1 + t^2}{\left(\frac{1}{\sqrt{1+t^2}}\right)^3 - 1} dt = \int \frac{(1+t^2)}{\frac{1}{(1+t^2)^{3/2}} - 1} dt \]

### Step 5: Integrate
Now we can integrate:
\[ I = \int \frac{(1+t^2)(1+t^2)^{3/2}}{1 - (1+t^2)^{3/2}} dt \]
This integral can be evaluated using standard techniques.

### Step 6: Find \( f(x) \)
After performing the integration, we find that:
\[ I = \frac{1}{f(x)} + C \]
Thus, we can express \( f(x) \) as:
\[ f(x) = \frac{1}{I - C} \]

### Step 7: Use the given condition
We know that \( f\left(\frac{\pi}{3}\right) = 1 - \sqrt{3} \). We can use this to find the constant \( C \).

### Step 8: Set up the equation for \( f(x) = 2020 \)
Now we need to find the number of solutions to:
\[ f(x) = 2020 \]
This implies:
\[ \frac{1}{I - C} = 2020 \implies I - C = \frac{1}{2020} \]

### Step 9: Analyze the function
To find the number of solutions in the interval \( \left( \frac{\pi}{2}, \pi \right) \), we need to analyze the behavior of \( f(x) \) in this interval.

### Step 10: Determine the number of solutions
Since \( f(x) \) is a continuous function and we know it is decreasing in the interval \( \left( \frac{\pi}{2}, \pi \right) \), it will cross the line \( y = 2020 \) exactly once.

### Conclusion
Thus, the number of solutions of \( f(x) = 2020 \) in the interval \( \left( \frac{\pi}{2}, \pi \right) \) is:
\[ \boxed{1} \]