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