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I have some problem with finding $\cos2x$ if $\cos^6x-\cos6x=1$

I replace the $\cos6x= \cos^23x-\sin^23x$ but I'am not sure it's a good way to figure it out.

Greor
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    This seems like an exercise that explores the Chebyshev polynomials of the first kind. http://en.wikipedia.org/wiki/Chebyshev_polynomials#Trigonometric_definition – Sammy Black Nov 20 '13 at 19:57
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    More to the point, you can write $\cos 6x$ as a $6$th-degree polynomial in the variable $\cos x$. Look here at $T_6(x)$. http://en.wikipedia.org/wiki/Chebyshev_polynomials#Examples – Sammy Black Nov 20 '13 at 19:59

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Your exercise is equivalent to finding $2u^2 - 1$, for $u$ a solution to the polynomial equation $$ u^6 - \left( 32u^6 - 48u^4 + 18u^2 - 1 \right) = 1. $$ This boils down to finding roots of the polynomial $$ 31u^6 - 48u^4 + 18u^2. $$

Sammy Black
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If $\displaystyle\cos2x=y,\cos6x=4y^3-3y$ as $\displaystyle\cos3A=4\cos^3A-3\cos A$

Again, $\displaystyle\cos^6x=\frac{(2\cos^2x)^3}8=\frac{(1+y)^3}8$

So, the equation becomes $$\frac{(1+y)^3}8-(4y^3-3y)=1$$

$$\iff -31y^3+3y^2+27y-7=0$$

lab bhattacharjee
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    @Greor, clearly one value of $y$ is $-1\implies \cos2x=-1=\cos\pi\implies 2x=(2n+1)\pi$ where $n$ is any integer. The rest two values can be easily deduced by solving the following quadratic equation : $$\frac{-31y^3+3y^2+27y-7}{y+1}=0$$ – lab bhattacharjee Nov 21 '13 at 04:11