问题描述: sin2x=sin(x-派/3)和cos(2x-派/6)=sin(3x-派/3)和sin5x=sin3x这种方程怎么解啊 1个回答 分类:数学 2014-11-21 问题解答: 我来补答 1.sin2x=sin(x-π/3) 则2x=2kπ+(x-π/3)或2x=π-(2kπ-π/3)+2kπ 所以x=2kπ-π/3 或x=4π/9+2kπ/32.cos(2x-π/6)=sin(3x-π/3) sin(π/2-(2x-π/6))=sin(3x-π/3) sin(2π/3-2x)=sin(3x-π/3) 则2π/3-2x=2kπ+(3x-π/3)或2π/3-2x=π-(3x-π/3)+2kπ 所以化简:x=(1-2kπ)/5或x=2kπ+π/33.5x=2kπ+3x或5x=π-3x+2kπ x=kπ或x=π/8+kπ/4 展开全文阅读