1is it 3?
we have 3x=60-2x
take cos on both sides
cos3x = cos(60-2x)
obtain a cubic equation in x which has 1 root as cos12
Find the minimum degree of a polynomial equation which has integral coefficients and has one of the roots as cos12.?
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6 Answers
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3mmm no it was given as 5.
but u r in the rite way,May i know y shud u take the eqn like this.
1i 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
3actually 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
341Wont 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