I am trying to solve the equation: 
solve(cos(3*x - Pi/12) = cos(2*x + Pi/7), x, allsolutions).
It is difficult for me to get the resullt. 
Solve by my hand. I tried
restart;
a := 3*x - Pi/12;
b := 2*x + Pi/7;
S1 := solve(a = 2*Pi*k + b, x);
S2 := solve(a = 2*Pi*k - b, x);
S := ((S1 union S2) assuming k::integer);
I get the answer

Can I get the result by command 
solve(cos(3*x - Pi/12) = cos(2*x + Pi/7), x, allsolutions).

I am trying to find three integer a, b, c so that equation (x-a)^4 + (x-b)^4 = c has four different solutions integer x. I tried

restart;
k := 0;
for a from -10 to 10 do
    for b from -10 to 10 do
for c from -10 to 10 do
if a*b*c <> 0 and igcd(a, b) = 1 then X := [solve((x - a)^4 + (x - b)^4 = c, x)]; if type(X[1], integer) and type(X[2], integer) and type(X[3], integer) and type(X[4], integer) and nops({X[1], X[2], X[3], X[4]}) = 4 then k := k + 1; L[k] := [a, b, c, X[]]; end if; end if; end do; end do;
end do;
L := convert(L, list);
k;
L;

I do not get any solution

I have a polyhedron with A(1,1,0), B(-1,1,0), C(-1,-1,0), H(-1/2,0,1), K(1/2,-1/2,0). How can I calculate volume of polyhedron ABCHK?
 

I want to put one option of tetrahedron SABC with AB=c, BC=a, AC=b, SA=m, SB=n, SC=p. I tried
restart;
with(Student:-MultivariateCalculus);
A := [0, 0, 0];
B := [c, 0, 0];
C := [(-a^2 + b^2 + c^2)/(2*c), sqrt((a + b - c)*(a - b + c)*(-a + b + c)*(a + b + c))/(2*c), 0];
S := [x, y, z];
solve([distance(S, A) = m, distance(S, B) = n, distance(S, C) = p], [x, y, z]);

I got warning, solutions may have been lost. How can I solve that system of equations?

I am trying to find remainder of (2 + sqrt(3))^15 + (2 - sqrt(3))^15 with 2017. I tried
irem((2 + sqrt(3))^15 + (2 - sqrt(3))^15, 2017);

I do not get the result. How can I get the result?

I use like this 
a := expand((2 + sqrt(3))^15 + (2 - sqrt(3))^15);
irem(a, 2017);

get the result.