14．函数f(x)=(2x−π)cos(π2−x)−sin(x−π2),x∈(−2π,3π)f(x)=(2x-\pi )\cos (\frac{\pi }{2}-x)-\sin (x-\frac{\pi }{2}),x\in (-2\pi ,3\pi )f(x)=(2x−π)cos(2π−x)−sin(x−2π),x∈(−2π,3π)的零点个数是((( )))