If $\sin A + \sin 2A = x$ and $\cos A + \cos 2A = y$ then the value of the expression $(x^2 + y^2)(x^2 + y^2 - 3)$ equals

Given equations are:

$x = \sin A + \sin 2A$

$y = \cos A + \cos 2A$

Squaring and adding both equations, we get:

$x^2 + y^2 = (\sin A + \sin 2A)^2 + (\cos A + \cos 2A)^2$

$x^2 + y^2 = (\sin^2 A + \cos^2 A) + (\sin^2 2A + \cos^2 2A) + 2(\sin A \sin 2A + \cos A \cos 2A)$

Using the identity $\cos(P - Q) = \cos P \cos Q + \sin P \sin Q$, we have:

$x^2 + y^2 = 1 + 1 + 2 \cos(2A - A)$

$x^2 + y^2 = 2 + 2 \cos A$

Now, substituting this value into the given expression $(x^2 + y^2)(x^2 + y^2 - 3)$:

$(2 + 2 \cos A)(2 + 2 \cos A - 3)$

$= (2 + 2 \cos A)(2 \cos A - 1)$

$= 4 \cos A - 2 + 4 \cos^2 A - 2 \cos A$

$= 4 \cos^2 A + 2 \cos A - 2$

$= 2(2 \cos^2 A - 1) + 2 \cos A$

Using the double angle identity $\cos 2A = 2 \cos^2 A - 1$, we get:

$= 2 \cos 2A + 2 \cos A$

$= 2(\cos A + \cos 2A)$

Since $y = \cos A + \cos 2A$, the expression simplifies to:

$= 2y$

Answer: $2y$