f(x)=cos^2x+cosxsinx(0＜x＜π/2) 1.若f(x)=1,求x的值 2,求f(x)的值域
问题描述：
f(x)=cos^2x+cosxsinx(0＜x＜π/2) 1.若f(x)=1,求x的值 2,求f(x)的值域
f(x)=cos^2x+cosxsinx(0＜x＜π/2)
1.若f(x)=1,求x的值
2,求f(x)的值域
答
f(x)=cos^2x+cosxsinx=(cos2x+1)/2+1/2sin2x=√2/2sin(2x+π/4)+1/2①f(x)=1 ,1=√2/2sin(2x+π/4)+1/2 ,√2/2=sin(2x+π/4) ,2x+π/4=3π/4 ,x=π/4②值域[-√2/2+1/2,√2/2+1/2]为何2x+π/4=3π/4而不是等于π/4?若2x+π/4=π/4时，x=0,而题目中(0＜x＜π/2),x≠0soga...谢谢.搞明白了