已知函数f(x)=2(sinx)^2+sin2x,x∈[0,2π],求使f(x)为正值的x的集合

题目

已知函数f(x)=2(sinx)^2+sin2x,x∈[0,2π],求使f(x)为正值的x的集合

答案

f(x)=2(sinx)^2+sin2x=1-cos2x+sin2x=√2sin(2x-π/4)+1
令f(x)＞0得
√2sin(2x-π/4)+1＞0
即sin(2x-π/4)＞-√2/2
因为x∈[0,2π]
所以-π/4≤2x-π/4≤15π/4
要满足sin(2x-π/4)＞-√2/2
则-π/4＜2x-π/4＜5π/4或7π/4＜2x-π/4＜13π/4
所以0＜x＜3π/4或π＜x＜7π/4
所以使f(x)为正值的x的集合是{x|0＜x＜3π/4或π＜x＜7π/4}