How do I introduce a new variable in a trigonometric equation?

Question

I have the trigonometric equation \begin{equation*} \sin^8 x + 2\cos^8 x -\dfrac{1}{2}\cos^2 2x + 4\sin^2 x= 0. \end{equation*} By putting $t = \cos 2x$, I have \begin{equation*} \dfrac{3}{16} t^4+ \dfrac{1}{4}t^3 + \dfrac{5}{8}t^2 -\dfrac{7}{4}t + \dfrac{35}{16} = 0. \end{equation*} How do I tell Mathematica to do that? Mathematica code is

Sin[x]^8 + 2 Cos[x]^8 - 1/2 Cos[2 x]^2 + 4 Sin[x]^2 == 0

Answer 1

You can use TrigExpand to expand all trigonometric functions to fundamental forms and then Eliminate solves the rest

eq1 = Sin[x]^8 + 2 Cos[x]^8 - 1/2 Cos[2 x]^2 + 4 Sin[x]^2 == 0;
eq2 = t == Cos[2 x]

Eliminate[TrigExpand[{eq1, eq2}], x]

Answer 2

One way to do this is:

Sin[x]^8 + 2 Cos[x]^8 - 1/2 Cos[2 x]^2 + 4 Sin[x]^2 == 0 /. 
  Solve[t == Cos[2 x], x] //FullSimplify // Expand // Union // Column // TraditionalForm

It gives exactly your answer if you get rid of your denominator 16 (multiply both sides of your equation by 16).

This will also work with more complex substitutions (for example t == Cos[x^2 - 1] ) when you can get multiple results:

Sin[x]^8 + 2 Cos[x]^8 - 1/2 Cos[2 x]^2 + 4 Sin[x]^2 == 0 /. 
Solve[t == Cos[x^2 - 1], x]//FullSimplify//Expand//Union //Column // TraditionalForm

Answer 3

A bit different approach :

Simplify @ TrigReduce[ Sin[x]^8 + 2 Cos[x]^8 - 1/2 Cos[2 x]^2 + 4 Sin[x]^2 == 0
/. Solve[ t == Cos[2 x], x, InverseFunctions -> True][[1]]]

35 + 10 t^2 + 4 t^3 + 3 t^4 == 28 t

or using Eliminate :

Eliminate[ TrigToExp[{ Sin[x]^8 + 2 Cos[x]^8 - 1/2 Cos[2 x]^2 + 4 Sin[x]^2 == 0,
                      t == Cos[2 x]}], x, InverseFunctions -> True] // 
PolynomialForm[#, TraditionalOrder -> True] &

3 t^4 + 4 t^3 + 10 t^2 - 28 t == -35  

Answer 4

First plot the trigonometric equation:

Plot[Sin[x]^8 + 2 Cos[x]^8 - 1/2 Cos[2 x]^2 + 4 Sin[x]^2 == 0, {x, 0, 2 Pi}]

You will see that there are no (real) solutions. Is this what you expect?