Question

设x属于(0,pai/2),则函数y=2sinx^2+1/sin2x的最小值为?

Answer 1

y=1-cos2x+1/sin2x.(x>pi/4 是增函数,X_min in (0,pi/4])

y'=2sin2x-2cos2x/sin^2 2x;

y'=0 iff sin^3 2x=cos 2x,sin^6 2x+sin^2 2x-1=0,2x in(0,pi/2].

求解(零点存在定理与单调性保证了存在唯一解) z^6+z^2-1=0,z in (0,1].

ymax=1+z^7+ z^5