已知函数f(x)满足f (cosx)=1/2 (x∈[0,π]),则f(-1/2)=?已知函数f(x)满足f (cosx)=1/2(x∈[0,π]),则f(-1/2)=?紧急紧急!

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已知函数f(x)满足f (cosx)=1/2 (x∈[0,π]),则f(-1/2)=?已知函数f(x)满足f (cosx)=1/2(x∈[0,π]),则f(-1/2)=?紧急紧急!

已知函数f(x)满足f (cosx)=1/2 (x∈[0,π]),则f(-1/2)=?已知函数f(x)满足f (cosx)=1/2(x∈[0,π]),则f(-1/2)=?紧急紧急!
已知函数f(x)满足f (cosx)=1/2 (x∈[0,π]),则f(-1/2)=?
已知函数f(x)满足f (cosx)=1/2(x∈[0,π]),则f(-1/2)=?紧急紧急!

已知函数f(x)满足f (cosx)=1/2 (x∈[0,π]),则f(-1/2)=?已知函数f(x)满足f (cosx)=1/2(x∈[0,π]),则f(-1/2)=?紧急紧急!
取x=2/3π, 满足x∈[0,π],则f(cosx)=1/2,又此时cosx=-1/2,f(cosx)=f(-1/2)=1/2

x∈[0,π], -->cosx∈[1,-1] 所以f(x)=1/2 x∈[1,-1] 所以f(-1/2)=1/2;