There is a trick to solving inequalities: convert them to equalities.
P/R < 1 implies P < R implies P + dP == R for some positive dP. That is, if you had to add something positive to P to get R, then it implies that P is strictly less than R.
(P+dP)==R implies P/R + dP/R == 1, which implies P/R == 1 - dP/R . So particular dP answers the question of how much less than one P/R is, proportional to R.
Q = @(v) sym(v,'r');
p = Q(1000);
L = Q(10);
a = Q(5);
B = Q(10);
W = Q(100000);
V_w = a*L*B+(((L^2)*B)/2);
V = B*(L*cos(x)*a+((L^2).*cos(x).*sin(x))/2+((L*cos(x))^2)/2);
theta_max = solve(V + dV == V_w, x, 'MaxDegree', 4);
theta_maxt = simplify( rewrite(theta_max, 'atan') );
The result you get out in theta_maxt will be a series of four long expressions involving atan. Each one of them answers the question of "What would the x have to be for V + dP to exactly equal V_w ?". Or in terms of the constants, "Given an exact amount less than 1000, what would x be for V to be exactly 1000 minus that amount?"
Now, depending on the difference from 1000 that you choose, some of the four theta_maxt might be imaginary. They involved roots of a quartic (degree 4 polynomial), so either 0, 2, or 4 of the solutions will be real valued. There are breakpoints at approximately dV = 839 point something, 1000 exactly, and 1361 point something (yes, there are solutions where x is negative.) The range between 839 and 1000 has four real-valued solutions.
When I put the original inequality to Maple as an inequality, Maple responds with four solutions, all involving arctan() of a particular constant, which I will call C for the moment. They are (arctan(C)-4*pi, 2*pi) exclusive, arctan(C)-2*pi,0) exclusive, (arctan(C),2*pi) exclusive, and (arctan(C)+2*pi,4*pi). The arctan(C) value is approximately 0.57 .