Question

高一三角函数，急！！！！

我在已知函数y=f(x)的定义域是[0,1/4],求下列函数的定义域：
1.f(sin^x)
2.f(cos^2-1/2）
参考书上关于第一题的答案是{x|2kπ-π/6≤x≤2kπ+π/6}或{2kπ+5π/6≤x≤2kπ+7π/6}
即【kπ-π/6,kπ+π/6】,k属于Z.
我想知道为什么{x|2kπ-π/6≤x≤2kπ+π/6}或{2kπ+5π/6≤x≤2kπ+7π/6}会等于【kπ-π/6,kπ+π/6】。

Answer 1

(1) 0 ≤ (cosx)^2 ≤ 1/4
-1/2 ≤ cosx ≤ 1/2
π/3 ≤x ≤ 2π/3 (cos(π/3) = 1/2, cos(2π/3) = -1/2)

(2) 0 ≤ (sinx)^2-1/2 ≤ 1/4
1/2 ≤ (sinx)^2 ≤ 3/4
√2/2 ≤ sinx ≤ √3/2 或 -√3/2 ≤ sinx ≤ -√2/2
π/4 ≤ x ≤ π/3 或 -π/3 ≤ x ≤ -π/4