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$\sqrt3\sin3x -\cos x =\sqrt2$

Does $\sin3x = 3\sin x - 4\sin^3x$ work?

$=3\sin x\cos^2x - \sin^3x$ ???

I don't see any factor and next step?

Solitarie
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    I would expand the $\sin 3x$ into $\sin x$ and $\cos x$, then use $\sin x = \pm \sqrt {1-\cos^2x}$ That will get you a cubic. You can hope the rational roots theorem solves your problem. In class problems it often does. If not, you can use Cardano and get a mess, or use a numeric root finder and get and approximate answer. – Ross Millikan Jan 05 '21 at 05:23
  • For what it's worth, Wolfram doesn't come up with any nice solutions: https://www.wolframalpha.com/input/?i=solve+sqrt%283%29+sin%283x%29-cos%28x%29%3Dsqrt%282%29. (It does come up with solutions, but only as particular roots of a degree-12 polynomial and that's not helpful.) – Semiclassical Jan 05 '21 at 05:43
  • You can get a sense of where the solutions are by plotting $\sqrt{3} \sin 3x$ and $\sqrt{2}+\cos x$ on the same graph: https://www.wolframalpha.com/input/?i=plot+sqrt%283%29+sin%283x%29+and+sqrt%282%29%2Bcos%28x%29 –  Jan 05 '21 at 06:37
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    I very strongly believe that the equation has to be divided by 2 on both sides and get nice ratios of $\frac{\sqrt{3}}{2}, \frac{1}{2}, \frac{1}{\sqrt{2}}$ –  Jan 05 '21 at 07:08

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