Question

已知θ∈(0,π/2),求f(x)=1/sin^2θ+2/cos^2θ的最小值

Answer 1

f(x)=1/sin^2θ+2/cos^2θ=1/(sinx)^2+2/(1-(sinx)^2)，令(sinx)^2=t,则t的取值范围为0

对f(t)求导有：

f'(t)=[(t+1)/(t-t^2)]'=[t-t^2-(t+1)(1-2t)]/(t-t^2)^2=(t^2+2t-1)/(t-t^...