3cos(2x+π/3)-1=0,x属于【0,2π】的解集
1个回答

由条件可以知道cos(2x+π/3)=1/3,

所以解得2x+π/3=arccos1/3+2kπ或 -arccos1/3+2kπ

即x=0.5arccos1/3+kπ -π/6或 -0.5arccos1/3+kπ -π/6

而x属于[0,2π],

故2x+π/3属于[π/3,13π/3]

得到解集为:

0.5arccos1/3 -π/6,0.5arccos1/3+ 5π/6,-0.5arccos1/3+ 5π/6,-0.5arccos1/3+ 11π/6,

-0.5arccos1/3+ 13π/6