Question

函数y=sin(2x+π/6)-cos(2x+π/3)的单调增区间为？

Answer 1

y=sin(2x+π/6)-cos(2x+π/3)
=sin2x*(√3/2)+cos2x*(1/2)-cos2x*(1/2)+sin2x*(√3/2)
=√3sin2x
则单调增区间为
-π+2kπ≤2x≤2kπ
-π/2+kπ≤x≤kπ,
即为[-π/2+kπ,kπ],7,把上式展开,可得根号3乘以sin2x
所以单调区间为:[-π/4,π/4]为增区间,[π/4.3π/4]为减区间,2,经过化简可以的y=跟号3sin2x，由图可知单调增区间-派/4+k派到派/4+k派，取闭区间,0,
y=sin(2x+π/6)-cos(2x+π/3)
=sin(2x+π/6)+sin(2x-π/6)
=2sin2xcosπ/6
=√3sin2x，
则单调增区间为
-π/2+2kπ≤2x≤π/2+2kπ
-π/4+kπ≤x≤π/4+kπ，
即为[-π/4+kπ,π/4+kπ]。
O(∩_∩)O~,0,