Answer to Question #143763 in Algebra for 1805040444b

Question #143763
Find at least one solution to the following equation:
sin(x^2 − 1)/1 − sin(x^2 − 1) = sin(x) + sin^2(x) + sin^3(x) + sin^4(x) + · · ·
1
Expert's answer
2020-11-17T06:08:38-0500

Right side of the equation is a geometric series with a=sin(x)a = sin(x) and r=sin2(x)sin(x)=sin(x)r = \frac{sin^2(x)}{sin(x)} = sin(x).

The sum of a geometric series:


S=a1rS = \frac{a}{1-r}S=sin(x)1sin(x)S = \frac{sin(x)}{1-sin(x)}

Hence the equation is:


sin(x21)1sin(x21)=sin(x)1sin(x)\frac{sin(x^2-1)}{1-sin(x^2-1)} = \frac{sin(x)}{1-sin(x)}

sin(x21)(1sin(x))=sin(x)(1sin(x21))sin(x^2-1)(1-sin(x)) = sin(x)(1-sin(x^2-1))

sin(x21)=sin(x)sin(x^2-1) = sin(x)

Since the period the function y=sin(x)y = sin(x) is 2π2\pi, the equation is:



x21=x+2πnx^2-1 = x +2\pi n

x2x(1+2πn)=0x^2-x-(1+2\pi n) = 0

x1=15+8πn2,x2=1+5+8πn2,n=0,1,2,...x_1 = \frac{1-\sqrt{5+8\pi n }}{2}, x_2 = \frac{1+\sqrt{5+8\pi n }}{2}, n = 0, 1, 2, ...


In particular, for n = 0 we have:


x1=152,x2=1+52x_1 = \frac{1-\sqrt{5}}{2}, x_2 = \frac{1+\sqrt{5}}{2}

For n = 1:


x1=15+8π2,x2=1+5+8π2x_1 = \frac{1-\sqrt{5+8\pi}}{2}, x_2 = \frac{1+\sqrt{5+8\pi}}{2}


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