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.