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This is one more of my unsolved trigonometry questions:

solve ;$$\sqrt{\frac{1-4\cos^2 4x}{8\cos (2x-2\pi/3)}}=\cos (2x-\pi/6)$$

My Try

provided that $\cos (2x-2\pi/3)\neq 0$ and squaring both sides $$1-4\cos^2 4x=8\cos (2x-2\pi/3)\cos^2(2x-\pi/6)....(1)$$ $$1-4\cos^2 4x=4\cos(2x-2\pi/3)(1+\cos(4x-\pi/3))...(2)$$

For convenience we take $2x=t$:$$1-4\cos^2 2t=4\cos(t-2\pi/3)(1+\cos(2t-\pi/3))....(3)$$

$$=4\cos(t-2\pi/3)-2\cos(3t)+2\cos (t+\pi/3)....(4)$$

What do i do next?

Source:A Panchishkin- Trigonometric functions

Albus Dumbledore
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  • I have to give up. I tried various ideas and I can't crack it. – user2661923 Oct 19 '20 at 20:38
  • @user2661923 no worries, i am still grateful to the fact that you tried,hopefully someone will comeup with a solution – Albus Dumbledore Oct 20 '20 at 03:03
  • Are you sure that there is no typo in the problem? Failed ideas: integration, differentiation, converting $(2x - \pi/6) \to$ $(2x - 2\pi/3)$, converting all to $g(x)$, where $g(x) \to 2x, 4x, (2x - 2\pi/3)$. Also, I tortured $\cos(a+b) = \cdots$ to death. It occurs to me that if you get lucky with some approach yielding one satisfying value for $x$, that might be sufficient if you can show, re $f'(x)$ et al, that the problem can only have one solution. – user2661923 Oct 20 '20 at 03:11
  • @user2661923 hmm i see you have tried more methods than me.Something intersting about the problem is that theanswer given is $k\pi$ that means we may have to factorise $\sinx$ and then show the rest of polynomial has no roots. I don't think there is a typo in the problem. – Albus Dumbledore Oct 20 '20 at 04:11

1 Answers1

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It's been a tour de force, mate. Even with the help of Wolfram Mathematica.

After expanding with standard formulae and subbed $2x=u$ I got this mess $$-2 \sin ^4 u-\sqrt{3} \sin u-2 \cos ^4 u+2 \cos ^3 u+\cos u+12 \sin ^2 u \cos ^2 u-6 \sin ^2 u \cos u-1=0$$ I substituted $$\sin u=\frac{2t}{1+t^2};\;\cos u=\frac{1-t^2}{1+t^2};\;t=\tan u/2$$

After a while I got $$t \left(3 t^7+\sqrt{3} t^6-39 t^5+3 \sqrt{3} t^4+73 t^3+3 \sqrt{3} t^2-13 t+\sqrt{3}\right)=0$$ And thanks to Mathematica this amazing factorization, using the function

Factor[t (Sqrt[3] - 13 t + 3 Sqrt[3] t^2 + 73 t^3 + 3 Sqrt[3] t^4 - 
    39 t^5 + Sqrt[3] t^6 + 3 t^7), Extension -> Automatic]

$$\left(\sqrt{3}-t\right) t \left(t+\sqrt{3}-2\right) \left(t+\sqrt{3}+2\right) \left(3 t+\sqrt{3}\right) \left(-3 t^3+3 \sqrt{3} t^2+9 t-\sqrt{3}\right)=0$$ The linear factors are trivial. The cubic is another mess. I got these three approximate solutions

$t_1=-1.19175, t_2=0.176327, t_3=2.74748$

A little consolation is that $t=\tan x$

In $[0,2\pi]$ solutions are $$x=0,\frac{5 \pi }{6},\pi ,\frac{4 \pi }{3},\frac{11 \pi }{6},2 \pi,0.174533,2.26893,5.41052,1.22173$$

Raffaele
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