How do you solve #sin^2x-1=0# and find all exact general solutions?

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Answer 1 · 2017-02-08T09:23:59

#x = (n-1/2)pi# or #(2 n-1)pi/2#

#sin^2 x-1=0#

#(sin x -1)(sin x +1) =0#

#sin x -1 =0, sin x =1, x= 1/2 pi, 5/2 pi,...#

#sin x + 1 =0, sin x =-1, x= 3/2 pi, 7/2 pi,...#

we can say that,

#x = 1/2 pi, 3/2 pi,5/2 pi, 7/2 pi,...#

we can use arithmetic progression to solve this sequence where,

#a=1/2 pi, d =pi#.

#T_n = 1/2 pi+(n-1)pi =(1/2+n-1)pi#

#T_n =(n-1/2)pi# or #(2 n-1)pi/2#

Therefore it general solution for #x# is #(n-1/2)pi# or #(2 n-1)pi/2# for all #n#