Question

函数y=sinx+1/cosx+3的值域为?

Answer 1

y=sinx+1/cosx+3

sinx+1=ycosx+3y

sinx-ycosx=3y-1

√(1+y^2)sin(x+k)=3y-1

sin(x+k)=(3y-1)/√(1+y^2)

|sin(x+k)|≤1

所以

|(3y-1)/√(1+y^2)|≤1

4y^2-3y≤0

所以0≤y≤3/4