1.求y=sinxcosx/(sinx+cosx+1)的值域.2.已知f(x)是周期是2π的函数,当x属于0到2π时,f(x)=s_数学解答

1.求y=sinxcosx/(sinx+cosx+1)的值域.
2.已知f(x)是周期是2π的函数,当x属于0到2π时,f(x)=sin(x/2),则f(x)=0.5的解集是.

人气：186 ℃　时间：2020-06-19 08:46:42

解答

1.y=2sinxcosx/2(sinx+cosx+1) =[(sinx+cosx)^2-1]/2(sinx+cosx+1) =(sinx+cosx-1)/2 =(√2/2)sin(x+π/4)-1/2所以y=sinxcosx/(sinx+cosx+1)的值域为:[-√2/2-1/2,√2/2-1/2]2.当x属于0到2π时,f(x)=sin(x/2),...