一道数学题,3sin2x+3=2sin3(x-π/12)+8,求x
问题描述：
一道数学题,3sin2x+3=2sin3(x-π/12)+8,求x
3sin2x+3=2sin3(x-π/12)+8,求x
3sin2x+3=2sin3(x-(π/12))+8
答
-1≤sin2x≤1,-1≤sin3(x-(π/12)）≤1所以方程左边最大值为6,右边最小值为6,所以2x=2Kπ+π/2,x=Kπ+π/43(x-(π/12)）=2Kπ-π/23x-π/4=Kπ3（Kπ+π/4）-π/4=2Kπ3Kπ+π/2=2Kπ-π/2解得K=-1所以x=Kπ+π/4=-3...