Are there any alternatives for NIntegrate to calculate the area for which $f(x,y)<0$?

Clear["Global`*"]

RegionPlot[
 2 Cos[(3 x)/2] + Cos[1/2 (x - 2 y)] + Cos[x/2] (7 + 8 Cos[x + y]) < 0,
 {x, 0, Pi}, {y, 0, Pi},
 Frame -> True]

rgn = ImplicitRegion[{2 Cos[(3 x)/2] + Cos[1/2 (x - 2 y)] + 
       Cos[x/2] (7 + 8 Cos[x + y]) < 0 && 0 < x < Pi && 
     0 < y < Pi}, {x, y}];

Area[rgn, WorkingPrecision -> MachinePrecision]

(* 3.49805 *)

Area[rgn, WorkingPrecision -> 15]

(* 3.49805 *)

Let us name the function of interest f[x,y]:

f[x_, y_] := (8 Cos[x + y] + 7) Cos[x/2] + Cos[(x - 2 y)/2] + 2 Cos[3 x/2]

Attempt to ContourPlot to find the zero lines:

ContourPlot[f[x, y] == 0, {x, 0, Pi}, {y, 0, Pi}]

Simple numerical evaluation determines that the negative region is the inside of these two curves. Notice that the ContourPlot is quite rapid and has very clean lines. Interesting coincidence. Perhaps there exists an analytical solution to these curve lines?

sol = Solve[{f[x, y] == 0}]

This returns a list of 4 possible curves, while also stating that some solutions may be missing. By manual inspection (such as by using Plot), we can find that the 2nd and 4th solutions are of interest to us, so we shall label them:

upperCurve = y /. sol[[2]];
lowerCurve = y /. sol[[4]];
Plot[{upperCurve, lowerCurve}, {x, 0, Pi}, PlotRange -> {0, Pi}]

Checking the curves manually by plotting them against the original ContourPlot, we see that upperCurve matches the upper line for the whole domain, and that lowerCurve matches the lower line up until it reaches its minimum.

Find the minimum of the lowerCurve:

FindMinimum[{lowerCurve, 0 < x < Pi}, x, WorkingPrecision -> 25]

{1.872299341324760554288429*10^-8, {x -> 2.094395111754692173633430}}

The warning about a small imaginary part is of little concern here, but you can increase the WorkingPrecision and PrecisionGoal if you would like more digits.

Michael Seifert also pointed out that an exact form can be found for this solution by applying TrigFactor to f[x,y]:

TrigFactor[f[x,y]]

2 (Cos[x/2 - y/2] + 2 Cos[x/2 + y/2]) (2 Cos[x + y/2] + Cos[y/2]) == 0

The lower line happens to correspond to the second variable factor in this expression, and its minimum is found when y is set to 0 and solved.

Solve[{(2 Cos[x + y/2] + Cos[y/2]) == 0 /. y -> 0, 0 < x < Pi}, x]

{{ x -> 2 Pi/ 3 }}

Integrate the area below the 2nd curve over the whole domain minus the area under the 4th curve for 0 through 2.094...

NIntegrate[upperCurve, {x, 0, Pi}, WorkingPrecision -> 25] - 
 NIntegrate[lowerCurve, {x, 0, 2.09439511175469217363342977478168904781`25.},
   WorkingPrecision -> 25]

3.49805583366099845069196

Or with the exact form, we can see a slightly different answer:

NIntegrate[upperCurve, {x, 0, Pi}, WorkingPrecision -> 25] - 
 NIntegrate[lowerCurve, {x, 0, 2 Pi/3}, WorkingPrecision -> 25]

3.49805583366099836305434

While this method is not universally applicable, it does work for this function and is much faster than Area or direct application of NIntegrate and Boole for high precisions. As yarchik notes, you can swap NIntegrate for Integrate here to acquire an exact solution, though it takes a bit longer to evaluate.

Method ->"LocalAdaptive" evaluates without errormessage

NIntegrate[Boole[2 Cos[(3 x)/2] + Cos[1/2 (x - 2 y)] +Cos[x/2] (7 + 8 Cos[x + y]) < 0], {x, 0, Pi}, {y,0, Pi}, Method -> "LocalAdaptive"]
(*3.49818*)

A similar way is

reg=ImplicitRegion[
   2 Cos[(3 x)/2] + Cos[1/2 (x - 2 y)] + Cos[x/2] (7 + 8 Cos[x + y]) <
     0, {{x, 0, Pi}, {y, 0, Pi}}];
NIntegrate[1, {x, y} ∈ reg]

3.49485