Question

求函数f（x）=2cos^2x+3sinx在[-π/2,π/2]上的最值

Answer 1

由于:

f(x)=2cos^2(x)+3sinx

=2*[1-sin^2(x)]+3sinx

=-2sin^2(x)+3sinx+2

设T=sinX

则：f(x)=-2T^2+3T+2

=-2(T-3/4)^2+25/8

由于X属于[-π/2,π/2]

则:T=sinX属于[-1,1]

则当T=3/4时,

f(x)有最大值=25/8

当T=-1时,

f(x)有最小值=-3