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Opalg · #5 October 16th, 2008, 04:07 AM

Use the formulas $\cos(2\theta) = 1 - 2\sin^2\theta$ and $\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$ . These show that if $x=4\sin^2\theta$ then $\cos(2\theta) = 1 - \tfrac x2$ , and so $\cos(6\theta) = 4\left(1 - \tfrac x2\right)^3 - 3\left(1 - \tfrac x2\right) = 1 - \tfrac92x + 3x^2 - \tfrac12x^3$ . But if $\theta = \tfrac\pi{18}$ then $\cos(6\theta)=\tfrac12$ . That gives the equation $x^3-6x^2+9x-1=0$ . The other two solutions will come from the other angles satisfying $\cos(6\theta)=\tfrac12$ , namely $\theta = \tfrac{5\pi}{18}$ and $\theta = \tfrac{7\pi}{18}$ . (ThePerfectHacker had $\theta = \tfrac{4\pi}{18}$ , but I think $\theta = \tfrac{5\pi}{18}$ must be correct.)

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