x^35=2
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MathBot Answer:
The solutions are given by the formula: \[x = \sqrt[35]{2} \cos{\left(\frac{2 \pi k}{35} \right)} + \sqrt[35]{2} \sin{\left(\frac{2 \pi k}{35} \right)} i\] where \(k\) is an integer in \(0 \le k < 35\).
There is \(1\) real solution.There are \(34\) solutions with nonzero imaginary part.
Real solutions: \[\begin{aligned}x &= \sqrt[35]{2} \\&\approx 1.0200016\end{aligned}\]
Solutions with nonzero imaginary part (\(9\) of \(34\) displayed): \[\begin{aligned}x &= \sqrt[35]{2} \cos{\left(\frac{2 \pi}{35} \right)} + \sqrt[35]{2} i \sin{\left(\frac{2 \pi}{35} \right)} \\&\approx 1.0036098 + 0.18212832 i\\x &= \sqrt[35]{2} \cos{\left(\frac{4 \pi}{35} \right)} + \sqrt[35]{2} i \sin{\left(\frac{4 \pi}{35} \right)} \\&\approx 0.95496107 + 0.35840289 i\\x &= \sqrt[35]{2} \cos{\left(\frac{6 \pi}{35} \right)} + \sqrt[35]{2} i \sin{\left(\frac{6 \pi}{35} \right)} \\&\approx 0.87561915 + 0.52315809 i\\x &= \sqrt[35]{2} \cos{\left(\frac{8 \pi}{35} \right)} + \sqrt[35]{2} i \sin{\left(\frac{8 \pi}{35} \right)} \\&\approx 0.76813411 + 0.67109856 i\\x &= \sqrt[35]{2} \cos{\left(\frac{2 \pi}{7} \right)} + \sqrt[35]{2} i \sin{\left(\frac{2 \pi}{7} \right)} \\&\approx 0.6359606 + 0.79746937 i\\x &= \sqrt[35]{2} \cos{\left(\frac{12 \pi}{35} \right)} + \sqrt[35]{2} i \sin{\left(\frac{12 \pi}{35} \right)} \\&\approx 0.4833468 + 0.89820886 i\\x &= \sqrt[35]{2} \left(- \frac{1}{4} + \frac{\sqrt{5}}{4}\right) + \sqrt[35]{2} i \sqrt{\frac{\sqrt{5}}{8} + \frac{5}{8}} \\&\approx 0.31519783 + 0.97007918 i\\x &= \sqrt[35]{2} \cos{\left(\frac{16 \pi}{35} \right)} + \sqrt[35]{2} i \sin{\left(\frac{16 \pi}{35} \right)} \\&\approx 0.13691815 + 1.0107704 i\\x &= - \sqrt[35]{2} \cos{\left(\frac{17 \pi}{35} \right)} + \sqrt[35]{2} i \sin{\left(\frac{17 \pi}{35} \right)} \\&\approx -0.045762199 + 1.0189745 i\end{aligned}\]
\(i\) is the imaginary unit, defined as \(i^2 = -1\).