# How can I get exactly solutions of this trigonometric equation?

I am trying to solve this solution

Sin[2 x + Pi/12] == Cos[3 x - Pi/5], -2 Pi <= x <= Pi


I tried

FullSimplify@
DeleteDuplicates[
x /. Solve[{Sin[2 x + Pi/12] == Cos[3 x - Pi/5], -2 Pi <= x <= Pi}, {x}]]


I got

I work around

Clear["Global*"]
a = 2 x + Pi/12;
b = Pi/2 - (3 x - Pi/5);
f[k_] = SolveValues[{a == b + k  2  Pi , -2 Pi <= x <= Pi}, x, Reals]


list1 = k /. Solve[-(637/120) < k < 263/120, k, Integers]

Table[f[k], {k, list1}]


g[k_] = SolveValues[{a == Pi - b + k  2  Pi , -2 Pi <= x <= Pi}, x, Reals]


list1 = k /. Solve[-(73/120) < k < 107/120, k, Integers]
Table[g[k], {k, list1}]


Form two cases, I get eight solutions.

How can I get all exactly solutions without working around?

• Solve[Sin[2 x+Pi/12]==Cos[3 x-Pi/5]&&-2 Pi<=x<=Pi//TrigToExp,x] Commented May 30 at 7:19
• @vector: This is an answer, not a comment. Commented May 30 at 8:04

TrigFactor helps:

 Solve[TrigFactor[Sin[2  x + Pi/12] - Cos[3  x - Pi/5]] == 0 &&
x >= -2*Pi && x <= Pi, x, Reals]


{{x -> -((563 \[Pi])/300)}, {x -> -((443 \[Pi])/300)}, {x -> -(( 323 \[Pi])/300)}, {x -> -((203 \[Pi])/300)}, {x -> -((83 \[Pi])/ 300)}, {x -> -((13 \[Pi])/60)}, {x -> (37 \[Pi])/300}, {x -> ( 157 \[Pi])/300}, {x -> (277 \[Pi])/300}}

Given that we know these roots are rational multiples of $$\pi$$， we can convert the result of Root objects to normal expressions as

Solve[{Sin[2  x + Pi/12] == Cos[3  x - Pi/5], -2  Pi <= x <=
Pi}, {x}] /. a_ArcTan :> Pi  Rationalize[N@a/Pi]

(*{{x -> -((203 \[Pi])/300)}, {x -> -((83 \[Pi])/300)}, {x -> -((
13 \[Pi])/60)}, {x -> (37 \[Pi])/
300}, {x -> -((563 \[Pi])/300)}, {x -> (157 \[Pi])/
300}, {x -> -((443 \[Pi])/300)}, {x -> (277 \[Pi])/
300}, {x -> -((323 \[Pi])/300)}}*)
$$$$