Question

求函数y=2cos²x+2sinxcosx的最大值,并指出最大值时x值得集合

Answer 1

原式=cos2x+1+sin2x
=√2sin(2x+π/4)+1
最大值时，2x+π/4=π/2+2kπ
解得x=π/8+kπ k属于z

Answer 2

y=2cos²x+2sinxcosx
y=2cos²x-1+2sinxcosx+1
y=cos2x+sin2x+1
y=√2sin(2x+45°)+1
即当sin(2x+45°)=1时，
y有最大值
所以
2x+45°=π/2+2kπ
x=π/8+kπ (k为整数)

Answer 3

y=2cos²x+2sinxcosx
=1+cos2x+sin2x
=1+sin(π/2-2x)+sin2x
=1+2sinπ/4cos(2x-π/4)
=1+√2cos(2x-π/4)
当cos(2x-π/4)=1, 2x-π/4=2kπ, x=kπ+π/8时, y最大=1+√2