The number of solutions of 2 cos x = |sin x|, 0 £ x £ 4 p, is

The given equation is 2 cos x = |sin x|, where x is in the range [0, 4π].

Step 1: Divide the range into two intervals [0, 2π] and [2π, 4π] because the functions sin x and cos x are periodic with period 2π.

Step 2: Solve the equation in the interval [0, 2π].

In this interval, sin x is positive in the first and second quadrants (0 ≤ x ≤ π) and negative in the third and fourth quadrants (π < x ≤ 2π).

So, we have two cases to consider:

Case 1: 0 ≤ x ≤ π, sin x is positive, so the equation becomes 2 cos x = sin x. This simplifies to 2 cos x - sin x = 0, which can be written as √5 sin(x + α) = 0, where α = arctan(2). The solutions to this equation in the interval [0, π] are x = nπ, where n is an integer.

Case 2: π < x ≤ 2π, sin x is negative, so the equation becomes 2 cos x = -sin x. This simplifies to 2 cos x + sin x = 0, which can be written as √5 sin(x + β) = 0, where β = arctan(-2). The solutions to this equation in the interval [π, 2π] are x = nπ, where n is an integer.

Step 3: Solve the equation in the interval [2π, 4π].

This is the same as the interval [0, 2π] because of the periodicity of sin x and cos x, so the solutions are the same.

Step 4: Combine the solutions from the two intervals.

The solutions to the equation 2 cos x = |sin x| in the interval [0, 4π] are x = nπ, where n is an integer.

So, there are 4 solutions in the interval [0, 4π].