To find the number of solutions for the equation |tan(2x)| = sin(x) in the interval [0, π/2], we can analyze the behavior of each function separately.

First, let's consider the function |tan(2x)|. Since the absolute value function does not affect the number of solutions, we can ignore it for now and focus on the equation tan(2x) = sin(x).

To simplify this equation, we can use the identity sin(2x) = 2sin(x)cos(x). Substituting this identity into the equation, we get:

tan(2x) = sin(x)
tan(2x) = 2sin(x)cos(x)

Now, let's consider the behavior of each function individually.

1) tan(2x): This function has an infinite number of solutions because it is periodic with a period of π. In the interval [0, π/2], there are no vertical asymptotes or points where the tangent function is undefined.

2) sin(x): This function has a maximum of one solution in the interval [0, π/2]. This is because sin(x) is an increasing function in this interval, starting from sin(0) = 0 and reaching its maximum value of sin(π/2) = 1.

Now, let's analyze the equation tan(2x) = 2sin(x)cos(x) in the interval [0, π/2].

Since the function tan(2x) has an infinite number of solutions and sin(x) has at most one solution, there can be at most one solution to the equation in this interval.

Therefore, the number of solutions of the equation |tan(2x)| = sin(x) in the interval [0, π/2] is at most one.