Given that
#4\sin2(x-1)-2=1#
#4\sin2(x-1)=3#
#\sin2(x-1)=3/4#
Writing general solutions,
#2(x-1)=2k\pi+\sin^{-1}(3/4)\ or#
#\ \ 2(x-1)=2k\pi+\pi-\sin^{-1}(3/4)#
#x=k\pi+1+1/2\sin^{-1}(3/4)\ \ or#
#\ \ x=(2k+1)\pi/2+1-1/2\sin^{-1}(3/4)#
Where, #k# is any integer i.e. #k=0, \pm1, \pm2, \pm3, \ldots#
But it given that #-2\pi\le x\le 2\pi#
hence setting #k=-2, -1, 0, 1# in above first general solution, we get four desired values
#x=-2\pi+1+1/2\sin^{-1}(3/4), \ -\pi+1+1/2\sin^{-1}(3/4), 1+1/2\sin^{-1}(3/4)\ \ , \pi+1+1/2\sin^{-1}(3/4)#
or
#x=-4.859, -1.717, 1.424, 4.565#
Similarly, setting #k=-2, -1, 0, 1# in above second general solution, we get another four desired values
#x=-{3\pi}/2+1-1/2\sin^{-1}(3/4), -{\pi}/2+1-1/2\sin^{-1}(3/4), {\pi}/2+1-1/2\sin^{-1}(3/4)\ \ , {3\pi}/2+1-1/2\sin^{-1}(3/4)#
or
#x=-4.136, -0.995, 2.147, 5.288#
Thus we get total eight desired values of #x# which are approximately
#x=-4.859,-4.136, -1.717, -0.995, 1.424, 2.147, 4.565, 5.288#
