题目

解方程：(1)sin2x=cos3x (2)sin(2x+π/3)+sin(x-π/6)=0 (3)sin(x+π/6)+cos(x+π/6)=0

提问时间：2020-10-16

答案

(1)sin2x-cos3x=0sin2x-sin(π/2-3x)=02sin(π/4-x/2)cos(5x/2-π/4)=0x/2-π/4=kπ或5x/2-π/4=kπ+π/2x=2kπ+π/2或x=2kπ/5+3π/10 (2)sin(2x+π/3)=-sin(x-π/6)=sin(π/6-x)2x+π/3=2kπ+π/6-xx=2kπ/3-π/18...