y=cos^2(x)-3cosx+1值域x∈[-π/4,π/3]

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y=cos^2(x)-3cosx+1值域x∈[-π/4,π/3]y=cos^2(x)-3cosx+1值域x∈[-π/4,π/3]y=cos^2(x)-3cosx+1值域x∈[-π/4,π/3]y=co

y=cos^2(x)-3cosx+1值域x∈[-π/4,π/3]
y=cos^2(x)-3cosx+1值域
x∈[-π/4,π/3]

y=cos^2(x)-3cosx+1值域x∈[-π/4,π/3]
y
= cos²x - 3cosx + 1
= (cosx - 3/2)² - 5/4
因为 x∈[-π/4,π/3]
所以 1/2 ≤ cosx ≤ 1
所以 -1 ≤ cosx - 3/2 ≤ -1/2
所以 1/4 ≤ (cosx - 3/2)² ≤ 1
所以 -1 ≤ (cosx - 3/2)² - 5/4 ≤ -1/4
所以值域是 [-1 ,-1/4]