# Reduce: This system cannot be solved with the methods available to Reduce

I got the error message 'Reduce::nsmet: This system cannot be solved with the methods available to Reduce.'

Here is my code

neweqns = {Sin[a] + Sin[a - b] + Sin[a - c] == 0 &&
Sin[a - b] == Sin[b] + Sin[b - c] && Sin[a] + Sin[c] == Sin[b] &&
Sin[a - c] + Sin[b - c] == Sin[c]}

result = Reduce[neweqns && 0 <= a <= 2 π && 0 <= b <= 2 π &&
0 <= c <= 2 π, {a, b, c}, Reals]


Could anyone help? Thank you!

• You should explain what you are going to achive, what the symbols are inteded to describe and you shouldn't use illegible (for the most of users) signs. 'Text...' doesn't seem to generate problems and so it is a bad idea to include it into your code, it should be a minimal example instead. Feb 10, 2022 at 11:51
• @Artes Yes, thank you. I just edited the code. I hope it looks more straightforward now.
– M.K
Feb 10, 2022 at 11:57
• Notation [something] is incorrect without the head of a function. The system includes 4 equations and only 3 unknown variables and so it is an overdetermined system. Although there are solutions, they are easy to guess. Feb 10, 2022 at 14:04
• You cannot use function brackets ([ ]) in place of parentheses (( )). Then use TrigExpand on neweqns. Feb 10, 2022 at 14:15
• @BobHanlon I don't understand. which [] brackets, the one in Reduce? Thank you!
– M.K
Feb 10, 2022 at 14:52

Let us expand sines of differences, then Reduce can crack it.

newesteqns =  Sin[a] + Sin[a]*Cos[b] - Cos[a]*Sin[b] + Sin[a]*Cos[c] -
Cos[a]*Sin[c] == 0 &&  Sin[a]*Cos[b] - Cos[a]*Sin[b] ==
Sin[b] + Sin[b]*Cos[c] - Cos[a]*Sin[c] &&  Sin[a] + Sin[c] == Sin[b] &&
Sin[a]*Cos[c] - Cos[a]*Sin[c] + Sin[b]*Cos[c] - Cos[b]*Sin[c] ==
Sin[c];
result = Reduce[newesteqns && 0 <= a <= 2 \[Pi] && 0 <= b <= 2 \[Pi] &&
0 <= c <= 2 \[Pi], {a, b, c}, Reals]


a == \[Pi] && ((b == 0 && (c == 0 || c == \[Pi] || c == 2 \[Pi])) || (b == \[Pi] && (c == 0 || c == \[Pi] || c == 2 \[Pi])) || (b == 2 \[Pi] && (c == 0 || c == \[Pi] || c == 2 \[Pi])))) || (b == 0 && ((a == 0 && (c == 0 || c == \[Pi] || c == 2 \[Pi])) || (a == 2 \[Pi] && (c == 0 || c == \[Pi] || c == 2 \[Pi])))) || (b == \[Pi] && ((a == 0 && (c == 0 || c == \[Pi] || c == 2 \[Pi])) || (a == 2 \[Pi] && (c == 0 || c == \[Pi] || c == 2 \[Pi])))) || (b == 2 \[Pi] && ((a == 0 && (c == 0 || c == \[Pi] || c == 2 \[Pi])) || (a == 2 \[Pi] && (c == 0 || c == \[Pi] || c == 2 \[Pi]))))

• you can also use TrigExpand to do the expansion: Reduce[And @@ TrigExpand@neweqns && 0 <= a <= 2 \[Pi] && 0 <= b <= 2 \[Pi] && 0 <= c <= 2 \[Pi], {a, b, c}, Reals] (+1)
– kglr
Feb 10, 2022 at 14:08
• @kglr: Thank you for your valuable comment. Feb 10, 2022 at 14:41