Right side of the equation is a geometric series with a=sin(x) and r=sin(x)sin2(x)=sin(x).
The sum of a geometric series:
S=1−raS=1−sin(x)sin(x) Hence the equation is:
1−sin(x2−1)sin(x2−1)=1−sin(x)sin(x)
sin(x2−1)(1−sin(x))=sin(x)(1−sin(x2−1))
sin(x2−1)=sin(x) Since the period the function y=sin(x) is 2π, the equation is:
x2−1=x+2πn
x2−x−(1+2πn)=0
x1=21−5+8πn,x2=21+5+8πn,n=0,1,2,...
In particular, for n = 0 we have:
x1=21−5,x2=21+5 For n = 1:
x1=21−5+8π,x2=21+5+8π
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