2 sin〖2(x-π/3)=√3〗
=>sin〖(2x-π/3)=√3/2〗
Basic Acute Angle= sin^(-1)(√3/2)
= π/3
The first and the second quadrants are where Sine is positive.
So,
2x-π/3= π/3 ,(π-π/3),(2π+π/3),(2π+π-π/3),(2π+2π+π/3),(2π+2π+π-π/3),
-(π+π/3),-(2π-π/3),-(2π+π+π/3),-(2π+2π-π/3),-(2π+2π+π+π/3),
-(2π+2π+2π-π/3)
Therefore,
x= π/3 ,π/2 , 4π/3, 3π/2, 7π/3, 5π/2, -5π/6, -π, -11π/6, -2π, -17π/6, -3π (Answer)
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