Find all solutions sin^(2)[x/2]=sin^(2)x

Question

Find solutions between 0 and pi

Answer 1

Given sin² (x/2) = sin²x, we can use half-angle formulas and identities, 

sin x/2 = +/- √[(1 - cosx)/2]  and 
sin² x = 1 - cos² x

So, replacing this for sin x/2, we get

(1 - cos x)/2 = sin²x

(1 - cos x)/2 = 1 - cos² x

(1 - cos x)/2 = (1 - cos x)(1 + cos x)

1/2 = 1 + cos x

-1/2 = cos x

cos x = -1/2 at x = 2∏/3

We also need to consider x = 0, since sin 0 = 0, and therefore sin 0/2 = 0 as well. 

So, between 0 and pi, the solutions are {0, 2∏/3}

 

Answer 2

sin x = 2 sin(x/2) cos(x/2)

So, (sin x/2)^2 = (sin x)^2 leads to

(sin x/2)^2 [1 - 4cos^2(x/2)] = 0

sin x/2 = 0 => x = 0

1 - 4cos^2(x/2) = 0 => cos(x/2) = +/- (1/2)

In [0, pi],

x/2 = pi/3 => x = 2pi/3

Answer: x = 0, 2pi/3

Answer 3

(sin x/2)^2  =  (sin x)^2

sin x/2  =  sin x

x= ±2pi/3, 0