# Solve trig equation

I have a simple trig equation:

Clear[f, x]
f[x_] = Cos[2*x] - 3 Sin[x] + 2
Solve[f[x] == 0, x]


I received a very long a complicated reply that I do not understand.

{{x -> ConditionalExpression[\[Pi] -
ArcTan[(-3 + Sqrt)/(2 Sqrt[2 (-1 + 3/4 (-3 + Sqrt))])] +
2 \[Pi] C, C \[Element] Integers]}, {x ->
ConditionalExpression[
ArcTan[(-3 + Sqrt)/(2 Sqrt[2 (-1 + 3/4 (-3 + Sqrt))])] +
2 \[Pi] C, C \[Element] Integers]}, {x ->
ConditionalExpression[
ArcTan[-I Sqrt[1/2 (1 - 3/4 (-3 - Sqrt))],
1/4 (-3 - Sqrt)] + 2 \[Pi] C,
C \[Element] Integers]}, {x ->
ConditionalExpression[
ArcTan[I Sqrt[1/2 (1 - 3/4 (-3 - Sqrt))],
1/4 (-3 - Sqrt)] + 2 \[Pi] C, C \[Element] Integers]}}


What is wrong with my formulation of the question?

• Solve[]is used to solve polynomial equations. Try FindRoot! – Ulrich Neumann May 29 '18 at 11:17
• use Solve[f[x] == 0, x] /. C -> 0 or Assuming[{Element[C, Integers]}, Simplify@Solve[f[x] == 0, x]] to get one of infinitely many (periodic) solutions. – kglr May 29 '18 at 11:18
• Reduce[f[x] == 0, x] // ToRadicals – march May 30 '18 at 4:06

You can solve your problem using Solve or NSolve:

 Solve[{f[x] == 0, 0 < x < 2 Pi}, x] // N
(*{{x -> 0.756171}, {x -> 2.38542}} *)


or using FindRoot with initial condition

 FindRoot[f[x] == 0, {x, 0}]
(* {x -> 0.756171}*)

Clear[f, x];
f[x_] = Cos[2*x] - 3 Sin[x] + 2;

FunctionPeriod[f[x], x]

Solve[{Cos[2*x] - 3 Sin[x] + 2 == 0, 0 < x < 2 Pi}, x] // ToRadicals // Simplify
N[%]

Solve[(Solve[Cos[2*x] - 3 Sin[x] + 2 == 0, x] /. {Rule -> Equal, List -> Or}) &&
0 < x < 2 Pi, x] // RootReduce // ToRadicals // Normal
N[%] 