已知函数f(x)=log2(1-sinx)+log2(1+sinx)的定义域为【-π/3,-π/4】,试求f(x)的最值

f(x)=log2(1-sinx)+log2(1+sinx)=log2[(1-sinx) (1+sinx)]
= log2(cos2x)
定义域为【-π/3,-π/4】,
所以cosx∈[1/2,√2/2], cos2x∈[1/4,1/2],
∴log2(1/4)≤log2(cos2x) ≤log2(1/2)
即-2≤log2(cos2x) ≤-1,
∴f(x)的最大值是-1,最小值是-2