已知函数f（x）=2cos（2x+π／4）解方程：2cos（2x+π／4）=1

2cos(2x+π/4)=1

cos(2x+π/4)=0.5

2x+π/4=±π/3+2πk

其中k∈Z

2x=-π/4±π/3+2πk

x=-π/8±π/6+πk

方程解集

x₁=π/24+πk

x₂=-7π/24+πk

f(x) 周期为 2π/2 = π；

2cos(2x+π/4) = 1

cos(2x+π/4) = 1/2

2x + π/4 = ±π/3

x1 = π/24，x2 = -7π/24

解集为 ( kπ + π/24 )∪( kπ - 7π/24 )， k∈Z 。