Question

能做几题是几题.

Answer 1

f(x)=sin(2x-(π/6))+cos(2x-(π/3))

=sin(2x-(π/6))+cos((π/3)-2x)

=sin(2x-(π/6))+sin(2x+(π/6))

=sin2x

1

2kπ+(π/2)≤2x≤2kπ+(π/2)

2

横坐标变为原来的1/2

3