SOLVE.. 2 cos 3x - 1 = 0
2cos 3x-1=02cos 3x=1cos 3x=1/23x=cos⁻¹ 1/2 (cos⁻¹ (1/2)= π/3 + 2Kπ) and cos⁻¹(1/2)=5π/3+2kπ1)3x=π/3+2kπx=π/9+2/3(kπ) K∈Z; K=...-2,-1,0,1,2...2)3x=5π/3+2kπx=5π/9+(2/3)kπAnswer: x=π/9+2/3(kπ) U 5π/9+(2/3)kπ ; K∈Z