f(x) = 2 sin x | cos x |, x ∈ [0, 6π]

f(x) = \(\begin{cases}2\,sin\,x\,cos\,x,&x\,\in[0,\frac{\pi}{2}]\cup[\frac{3\pi}{2},\frac{5\pi}{2}]\cup[\frac{7\pi}{2},\frac{9\pi}{2}]\cup[\frac{11\pi}{2},6\pi]\\-2\,sin\,x\,cos\,x,&x\,\in[\frac{\pi}{2},\frac{3\pi}{2}]\cup[\frac{5\pi}{2},\frac{7\pi}{2}]\cup[\frac{9\pi}{2},\frac{11\pi}{2}]\end{cases}\)

⇒ f(x) = \(\begin{cases}sin\,2x,&x\,\in[0,\frac{\pi}{2}]\cup[\frac{3\pi}{2},\frac{5\pi}{2}]\cup[\frac{7\pi}{2},\frac{9\pi}{2}]\cup[\frac{11\pi}{2},6\pi]\\-sin\,2x,&x\,\in[\frac{\pi}{2},\frac{3\pi}{2}]\cup[\frac{5\pi}{2},\frac{7\pi}{2}]\cup[\frac{9\pi}{2},\frac{11\pi}{2}]\end{cases}\)

By differentiating f(x) w.r.t. x, we obtain,

f'(x) = \(\begin{cases}2\,cos\,2x,&x\,\in[0,\frac{\pi}{2}]\cup[\frac{3\pi}{2},\frac{5\pi}{2}]\cup[\frac{7\pi}{2},\frac{9\pi}{2}]\cup[\frac{11\pi}{2},6\pi]\\-2\,cos\,2x,&x\,\in[\frac{\pi}{2},\frac{3\pi}{2}]\cup[\frac{5\pi}{2},\frac{7\pi}{2}]\cup[\frac{9\pi}{2},\frac{11\pi}{2}]\end{cases}\)

For critical points, we have to find these x which gives f'(x) = 0

Now, f'(x) = 0 gives cos 2x = 0 = cos(2n+1)π/2

⇒ 2x = (2n+1)π/2

⇒ x = (2n+1)π/4, n ∈ 0, 1, 2, .... , 12.

Hence, number of critical points are 13.

Alternative :→ f(x) = 2 sin x | cos x |

⇒ f(x) = | sin 2x |

Then f'(x) = | 2 cos 2x | 

Thus, f'(x) = 0 gives | cos 2x | = 0

⇒ cos 2x = 0 (∵ |x| = 0 ⇒ x = 0)

⇒ cos 2x = 0 = cos(2n+1)π/2

⇒ 2x = (2n+1)π/2

⇒ x = (2n+1)π/4

Since, x ∈ [0, 6π]

∴ x = (2n+1)π/4, n = 0, 1, 2, .... , 12 are critical points of function f(x).

Hence, number of critical points = 13.