How do i solve this question?

#4 sin 2 (x-1)-2 =1# between the domains of #-2pi <= x <= 2 pi #

1 Answer

Answer:

See the answers below,

Explanation:

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#