Using an iterative process.
Making f(x,y) = ((4xy+4y^2-3cos(y)+2),(x^2-2xy^2-5sin(x)+5))
Calling p = (x,y) and p_0 = (x_0,y_0)
The linear expansion around x_0,y_0 is
f(p) approx f(p_0)+grad f_(p_0)(p-p_0)
Supposing that abs(f(p)) approx 0 we have the iterative procedure
p = p_0 - (grad f_(p_0))^(-1)f(p_0) or
p_(k+1) = p_k - (grad f_(p_k))^(-1)f(p_k)
Here
grad f = ((4 y, 4 x + 8 y + 3 sin(y)),(2 x - 2 y^2 - 5 cos(x), -4 x y))
Starting with p_0 = (2.32703,-2.11049) we obtain the convergent sequence
((1.70737,-1.42624),
(1.15774,-1.01564),
(1.00499,-0.939189),
(1.00591,-0.942905),
(1.00591,-0.942895))
within 6 significant digits.