作业帮设函数f(x)=cos(2x+π/3)+sin^2x-1/2,当x∈[0,π]时,f(x)的值域为
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![设函数f(x)=cos(2x+π/3)+sin^2x-1/2,当x∈[0,π]时,f(x)的值域为](/uploads/image/z/14794761-57-1.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Dcos%282x%2B%CF%80%2F3%29%2Bsin%5E2x-1%2F2%2C%E5%BD%93x%E2%88%88%5B0%2C%CF%80%5D%E6%97%B6%2Cf%28x%29%E7%9A%84%E5%80%BC%E5%9F%9F%E4%B8%BA)
设函数f(x)=cos(2x+π/3)+sin^2x-1/2,当x∈[0,π]时,f(x)的值域为
设函数f(x)=cos(2x+π/3)+sin^2x-1/2,当x∈[0,π]时,f(x)的值域为
设函数f(x)=cos(2x+π/3)+sin^2x-1/2,当x∈[0,π]时,f(x)的值域为
f(x)=cos(2x+π/3)+sin^2x-1/2=cos(2x+π/3)-1/2cos2x=-√3/2sin2x
当x∈[0,π]时,f(x)的值域为[-√3/2,√3/2]
(x)=cos(2x+π/3)+sin^2x-1/2
=cos2xcosπ/3-sin2xsinπ/3+(1-cos2x)/2-1/2
=1/2cos2x-√3/2*sin2x+1/2-1/2cos2x-1/2
=-√3/2sin2x
x∈[0,π]
2x∈[0,2π]
f(x)的递减区间:(0,π/2)U(3π/2,2π