sinx - sin3x/sin^2x -cos^2x =sin2x

Since you did not mentioned what do you want to do with
the given relation: to verify the identity or to solve the equation, I'll solve the
equation.

(sinx - sin3x)/[(sinx)^2 -(cosx)^2]
=sin2x

We'll apply the double angle identity for the
denominator:

cos 2x = (cosx)^2 -
(sinx)^2

We'll transform into a product the
numerator:

sinx - sin3x = 2 cos [(x+3x)/2]*sin[(x -
3x)/2]

sinx - sin3x = 2 cos 2x*sin (-
x)

Since the sine function is odd, we'll
have:

sin (- x) = - sin x

sinx
- sin3x = - 2*cos 2x*sin x

- 2*cos 2x*sin x/-cos 2x
=sin2x

We'll simplify and we'll
get:

2 sin x = sin 2x

2 sin x
- sin 2x = 0

2 sin x - 2 sin x* cos x =
0

We'll factorize by 2 sin x:

2 sin x(1 - cos x) = 0

We'll cancel each
factor:

2 sin x = 0

sin x =
0

x = (-1)^k arcsin 0 + k pi

x
= k*pi

1 - cos x = 0

cos x =
1

x = + arccos 1 + 2k*pi

x =
2k*pi

The solutions of the equation are:
{k*pi}U{2k*pi}.