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How does one determine how close $\theta = \frac{\pi}{\sqrt{2}}$ is to being an explicit solution of the equation $\cos \theta = -\left(\frac{1 + \sqrt{7}}{6 }\right)$?

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  • $\begingroup$ Are you asking merely for the existence of $\theta$, or what the exact, actual value of $\theta$ is? I don't think you can get the actual value in this case. $\endgroup$ – астон вілла олоф мэллбэрг Oct 31 '17 at 4:50
  • $\begingroup$ The existence of some $\theta$ is clear since the value on the right hand side of the equation lies in the range of $\cos(\cdot)$. Actually I was hoping to find $\theta$ as a fraction of $2\pi$. $\endgroup$ – Anton Van Wyk Oct 31 '17 at 5:05
  • $\begingroup$ No, unfortunately I do not think that is the case. $\endgroup$ – астон вілла олоф мэллбэрг Oct 31 '17 at 5:07
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It looks like it's not a rational multiple of $\pi$, so it's likely that the best closed-form answer is $$\theta = \cos^{-1}\left(\frac{-1-\sqrt{7}}{6}\right)$$

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  • $\begingroup$ I agree and numerical experimentation suggests that one such solution is approximately $\frac{\pi}{\sqrt{2}}$. How does one prove or disprove $\theta = \frac{\pi}{\sqrt{2}}$ as a possible solution? $\endgroup$ – Anton Van Wyk Oct 31 '17 at 5:13
  • $\begingroup$ It is pretty close, but the cosines differ by about 0.001. Straightforward calculation might be your best bet? $\endgroup$ – Andrew Tindall Oct 31 '17 at 5:19
  • $\begingroup$ If you really need to do it with pencil and paper, you can do a Taylor Series for $\cos(\pi/\sqrt{2})$ and a continued fraction for $(-1 - \sqrt{7})/6$ :) $\endgroup$ – Andrew Tindall Oct 31 '17 at 5:22

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