minimum degree

mmm no it was given as 5.

but u r in the rite way,May i know y shud u take the eqn like this.

i forgot the condition of integral coefficients in my previous post
we have 4cos^3x-3cosx=cos60.cosx-sin60.sinx
or 4cos^3x-3cosx=\frac{cosx}{2}-\frac{\sqrt{3}sinx}{2}

\frac{\sqrt{3(1-cos^2x)}}{2}=\frac{cosx}{2}+3cosx-4cos^3x

3(1-cos^2x)=(7cosx-8cos^3x)^2
so i think it should be 6

actually in the qn it says only one of the roots is cos 12' y dont we have an eqn whose roots are greater than 1

Wont cos 5x = 1/2 do the trick?

Since cos 5x + cos x = 2 cos 3x cos 2x

we have cos 5x = 2 cos 3x cos 2x - cos x = 1/2. Multiply both sides by 2 and you have a quintic in cos x

32 cos⁵x-40 cos³x+10 cosx -1 = 0

so that cos 12 is a root of 32t⁵-40t³+10 t-1 = 0