toandhsp

3 years, 244 days


These are questions asked by toandhsp

I want to reduce all solution of the equation sin(x)^2=1/4

restart:
sol:=solve(sin(x)^2=1/4, x, AllSolutions);

and

restart:
k:=combine((sin(x))^2);
sol:=solve(k=1/4, x, AllSolutions = true, explicit);
simplify(sol);

How can I reduce solution sol := -1/3*Pi*_B3+1/6*Pi+Pi*_Z3 ?

How can I get x= pi/6+k*pi and x= -pi/6+k*pi?

How can I solve this equation 18 *9^(x^2 + 2* x) + 768* 4^((x + 3)* (x - 1)) - 5 *6 ^((x + 1)^2)?

I tried

restart:

A:=18 *9^(x^2 + 2* x) + 768* 4^((x + 3)* (x - 1)) - 5 *6 ^((x + 1)^2);
solve(A=0);

I see that, the equation has three solutions: x = -2, x = -1 and x = 0. I check

f:=x->18 *9^(x^2 + 2* x) + 768* 4^((x + 3)* (x - 1)) - 5 *6 ^((x + 1)^2);

f(-2);

f(-1);

f(0);

Another question, Maple can not solve inequality 

18 *9^(x^2 + 2* x) + 768* 4^((x + 3)* (x - 1)) - 5 *6 ^((x + 1)^2) > = 0.

PS. We can easy to solve the above inequality with Mathematica

Reduce[18 9^(x^2 + 2 x ) + 768 4^((x + 3) (x - 1)) - 5 6 ^((x + 1)^2) >= 0 , x, Reals]

I got x <= -2 || x == -1 || x >= 0

 

 

When I use fsolve with equation 

-x^2 + 2*x + 5 + (x^2 + 2*x - 1)* sqrt(2 - x^2)=0

I got only one solution.

fsolve(-x^2 + 2*x + 5 + (x^2 + 2*x - 1)* sqrt(2 - x^2)=0,x);

In fact, it have two reals solutions.  

I posted at here

http://mathematica.stackexchange.com/questions/83985/does-the-equation-have-two-roots/83991#83991

 

I have a list L:={1,2,3,4,5,6,7,8}. I choose three elements from list L. How many different combinations of three numbers can be selected from L so that the numbers could represent the side lengths of a triangle?

I tried

nops({x, y, z,1+x, 2+y, -3+z,-2+x, 3+y, -1+z });

and Maple out 9.

If x = 1, y = 2, z = 3, we have

nops({1, 2, 3, 2, 4, 0, -1, 5, 2});

equal to 7.

Edit.

I edited my question. The word "list" into "set".

My question means, there is a triple (x, y, z) = (1, 2, 3) so that the number of element of the set {x, y, z,1+x, 2+y, -3+z,-2+x, 3+y, -1+z } is 7, not 9. Why we can confirm the number of elements of the set {x, y, z, 1+x, 2+y, -3+z,-2+x, 3+y, -1+z } is 9? And, how must select the integer numbers x, y, z (0 < x <10, 0< y <10, 0< z < 10) so that the number of elements of the set {x, y, z, 1+x, 2+y, -3+z,-2+x, 3+y, -1+z } is 9?

I tried

restart:

ListTools[Categorize]:

L:=[]:

for x from -2 to 5  do

for y from -2 to 5  do

for z from -2 to 5  do

a:=[x,y,z]:

b:=[1+x,2+y,-3+z]:

c:=[-2+x,3+y,-1+z]:

if op(1,a)*op(2,a)*op(3,a)*op(1,b)*op(2,b)*op(3,b)*op(1,c)*op(2,c)*op(3,c)<> 0

and op(1,a)<>op(2,a) and  op(1,a)<>op(3,a) and  op(1,a)<>op(2,b) and  op(1,a)<>op(3,b) and  op(1,a)<>op(2,c) and  op(1,a)<>op(3,c) and op(2,a)<>op(3,a) and op(2,a)<>op(1,b) and

op(2,a)<>op(3,b) and

op(2,a)<>op(1,c) and

op(2,a)<>op(3,c) and

op(3,a)<>op(1,b) and

op(3,a)<>op(2,b) and 

op(3,a)<>op(1,c) and

op(3,a)<>op(2,c) and

op(1,b)<>op(2,b) and

op(1,b)<>op(3,b) and

op(1,b)<>op(2,c) and

op(1,b)<>op(3,c) and

op(2,b)<>op(3,b) and

op(2,b)<>op(1,c) and

op(2,b)<>op(3,c) and

op(3,b)<>op(1,c) and

op(3,b)<>op(2,c) and

op(1,c)<>op(2,c) and

op(1,c)<>op(3,c) and op(2,c)<>op(3,c) then L:=[op(L), {a,b,c}] fi; od: od: od:

nops(L); 

L;

 

 

 

 

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