求值域y=3+cos(2x-π/3)

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求值域y=3+cos(2x-π/3)求值域y=3+cos(2x-π/3)求值域y=3+cos(2x-π/3)解因2x-π/3∈R所以cos(2x-π/3)∈[-1,1]3+cos(2x-π/3)∈[2

求值域y=3+cos(2x-π/3)
求值域y=3+cos(2x-π/3)

求值域y=3+cos(2x-π/3)

因2x-π/3∈R
所以cos(2x-π/3) ∈[-1,1]
3+cos(2x-π/3) ∈[2,4]
即y∈[2,4]

-1≤cos≤1
2≤y≤4

多看看书,cos(),无论括号内为多少,其值域都是[-1,1]