In geometry 3D, I have four vertices A(x1,y1,z1), B(x2,y2,z2), C(x3,y3,z3) and D(x4,y4,z4) of a tetrahedron. How to check a point P(x0,y0,z0) inside, outside or lie on  tetrahedron ABCD?

This is a problem of Russian in 2005: Solve inequality 

(3-x-sqrt(5-x^2))/(cos((2*x-7)/4)-cos((x-5)/4))>=0.

The problem can solve by hand. We only solve

(3-x-sqrt(5-x^2))/((2*x-7)/4-(x-5)/4) >=0.

I tried

eq1:=(3-x-sqrt(5-x^2))/((2*x-7)/4-(x-5)/4);

solve(eq1>=0,x);

I got

RealRange(1,Open(2)), RealRange(Open(2),5^(1/2)).

If I solve directly by

eq:=(3-x-sqrt(5-x^2))/(cos((2*x-7)/4)-cos((x-5)/4));

solve(eq>=0,x);

I don't get the exact answer. 

P.S. with Mathematica, I only use

Reduce[(3 - x - Sqrt[5 - x^2])/(Cos[(2 x - 7)/4] - Cos[(x - 5)/4]) >=
0, x, Reals]

With Maple, I tried

solve(sin(Pi*(x+1)/(4*x^2-4*x+2))=cos(Pi*(x-2)/(4*x^2-4*x+2)),x);

And I got

Warning, solutions may have been lost

RootOf(-sin(1/2*Pi*(_Z+1)/(2*_Z^2-2*_Z+1))+cos(1/2*Pi*(_Z-2)/(2*_Z^2-2*_Z+1)))

With Mathematica

Reduce[Sin[(x + 1) \[Pi]/(4 x^2 - 4 x + 2)] == Cos[(x - 2) \[Pi]/(4 x^2 - 4 x + 2)], x, Reals]

I got all solutions.

I tried

solve(log[3+sqrt(8)](3-x)-log[17-6*sqrt(8)](4*x^2+20*x+25)+log[3-sqrt(8)](x^2+x-2)>=0,x);

and got 

Warning, solutions may have been lost

RealRange(-infinity,Open(-5/2)), RealRange(Open(-5/2),Open(-2)), RealRange(Open(1),Open(3))

How can I get the exact the solution?

I want to solve the inequality sqrt(8*16^x - 1/2* 9^x) <= 3*4^x - 3^x

I tried 

restart:

solve(sqrt(8*16^x - 1/2* 9^x) <= 3*4^x - 3^x,x);

And I got 

Error, (in evala/inv/heuristic) cannot allocate memory (size=117506048).

How to solve the above inequality?