Question

两道题，要过程，谢谢

Answer 1

3x=π/2±x+2kπ
1.3x=π/2+x+2kπ
则2x=π/2+2kπ
x=π/4+kπ
2.3x=π/2-x+2kπ
4x=π/2+2kπ
x=π/8+kπ/2
所以x=π/4+kπ或π/8+kπ/2
cos[π/2-(π/3+x)]=cos(π/6-x)=sin(π/3+x)
y=sin(x+π/3)cos(π/6-x)
=sin(x+π/3)sin(π/3+x)
=sin²(π/3+x)
=[1-cos(2π/3+2x)]/2
=1/2-1/2cos(2π/3+2x)
T=2π/2=π

Answer 2

2）
sin3x-cosx=0
sin3x-sin(x+π/2)=0
2sin(2x+π/4)cos(x-π/4)=0
sin(2x+π/4)=0, 或cos(x-π/4)=0
得2x+π/4=kπ, 或x-π/4=kπ+π/2
得x=kπ/2-π/8, 或x=kπ+3π/4
这里k为任意整数

5）y=1/2[sin(2x)+sin(2π/3)]
最小正周期T=2π/2=π

Answer 3

这黑提子露出来都吓跑了

Answer 4

好黑的手