If $\displaystyle\int \dfrac{\operatorname{cosec}^2 x - 2010}{\cos^{2010} x}\,dx = -\dfrac{f(x)}{(g(x))^{2010}} + c$, where $f\left(\dfrac{\pi}{4}\right) = 1$; then the number of solutions of the equation $\dfrac{f(x)}{g(x)} = \{x\}$ in $[0, 2\pi]$ is/are (where $\{\cdot\}$ represents fractional part function)

Let $I = \int \dfrac{\operatorname{cosec}^2 x - 2010}{\cos^{2010} x}\,dx$

We can rewrite the integral as:

$I = \int (\operatorname{cosec}^2 x - 2010) \cos^{-2010} x \,dx$

Consider the derivative of $\cot x \cos^{-2010} x$:

$\dfrac{d}{dx} \left( \cot x \cos^{-2010} x \right) = -\operatorname{cosec}^2 x \cos^{-2010} x + \cot x (-2010) \cos^{-2011} x (-\sin x)$

$= -\operatorname{cosec}^2 x \cos^{-2010} x + 2010 \left( \dfrac{\cos x}{\sin x} \right) \cos^{-2011} x \sin x$

$= -\operatorname{cosec}^2 x \cos^{-2010} x + 2010 \cos^{-2010} x$

$= -(\operatorname{cosec}^2 x - 2010) \cos^{-2010} x$

Therefore, the integral is:

$I = -\cot x \cos^{-2010} x + c = -\dfrac{\cot x}{\cos^{2010} x} + c$

Comparing this with the given expression $-\dfrac{f(x)}{(g(x))^{2010}} + c$, we get:

$f(x) = \cot x$ and $g(x) = \cos x$

We are given $f\left(\dfrac{\pi}{4}\right) = \cot\left(\dfrac{\pi}{4}\right) = 1$, which satisfies the condition.

Now, we need to find the number of solutions for the equation $\dfrac{f(x)}{g(x)} = \{x\}$ in $[0, 2\pi]$.

$\dfrac{f(x)}{g(x)} = \dfrac{\cot x}{\cos x} = \dfrac{1}{\sin x} = \operatorname{cosec} x$

The equation becomes $\operatorname{cosec} x = \{x\}$.

We know that the range of the fractional part function is $0 \le \{x\} < 1$.

However, the range of $\operatorname{cosec} x$ is $(-\infty, -1] \cup [1, \infty)$.

Since the ranges of the two functions do not intersect, there are no solutions to the equation $\operatorname{cosec} x = \{x\}$ in $[0, 2\pi]$.

Thus, the number of solutions is $0$.

Answer: $0$