求y=cos2x-3cos3x+2的值域
1个回答

y=cos2x-3cosx+2

=2cos^2 x-1-3cosx+2

=2cos^2 x-3cosx +1

=2(cos^2 x -3cosx/2)+1

=2(cos^2 x -3cosx/2 + 9/16)+1-9/8

=2(cosx-3/4)^2 -1/8

当cosx=3/4时将取得最小值-1/8

当cosx=-1时,将取得最大值2(-1-3/4)^2-1/8=6

所以值域是[-1/8,6]