A Curious Inequality with... Radicals // ... रेडिकलहरूसँगको एउटा जिज्ञासु असमानता

 Nu mai stiu

Dupa cum xice pe coperta

(to be continue)

Neculai STANCIU (Buzău) in the MATHEMATICAL JOURNAL // Neculai STANCIU (Buzău) în GAZETA MATEMATICĂ

 In the recent issue of the Exercise Supplement (aka SGM) he proposed the Problem S:E25.199. The problem statement is :

                    "Consider  $\Delta ABC$  a triangle such that  $\measuredangle A=2\measuredangle C$.

                      Prove that  $\frac{AB}{BC}=\frac{1}{2\cos C}.$

Neculai STANCIU, Buzău"

Solution CiP

We construct  $AD$ - the bisector of angle  $\widehat{BAC}$. We have 

$$\measuredangle CAD=\measuredangle BAD=\frac{\measuredangle BAC}{2}=\measuredangle C$$

          We now construct  $BE\parallel AD,\;E\in AC$. We have for the angles formed with the secant $AB$  that

$\measuredangle BAD=\measuredangle ABE=\measuredangle C \tag{1}$

and the exterior angle  $\widehat{BAC}$ of triangle  $\Delta ABE$ shows us that

$\measuredangle BAC=\measuredangle ABE+\measuredangle AEB\Leftrightarrow 2\measuredangle C\overset{(1)}{=}\measuredangle C+\measuredangle AEB\Rightarrow \measuredangle AEB=\measuredangle C$

From all the angles equal to  $C$  in the figure, we see that we have the isosceles triangles  $BAE$  and  $CBE$, so

$ AB=AE\;,\;BC=BE\tag{2}$

Constructing the height  $AF$  in the isosceles triangle  $ABE,\;AF\perp BE$, it will also be the median, so  $BF=EF=BE/2$. Then, in triangle  $ABF$

$\cos C\underset{(1)}{=}\cos \widehat{ABF}=\frac{BF}{AB}=\frac{BE/2}{AB}\underset{(2)}{=}\frac{BC/2}{AB}=\frac{BC}{2\cdot AB}$,  which is equivalent to the statement  $\frac{AB}{BC}=\frac{1}{2\cos C}.$

$\blacksquare$

          Remark CiP  The solution does not use any advanced trigonometry formulas, only definitions, otherwise from the Law of Sines in triangle  $ABC$  we could easily obtain

$\frac{AB}{\sin C}=\frac{BC}{\sin A}\Rightarrow \frac{AB}{BC}=\frac{\sin C}{\sin A}=\frac{\sin C}{\sin 2C}=\frac{\sin C}{2\sin C \cdot \cos C}=\frac{1}{2\cos C}$.

Starting from a Trigonometric Identity without variables // Ξεκινώντας από μια Τριγωνομετρική Ταυτότητα χωρίς μεταβλητές

           We have on my Page, at No. 6, the Identity

$\frac{\tan10^{\circ}}{\tan20^{\circ}\cdot \tan40^{\circ}}=\frac{1}{\sqrt{3}}$

It is equivalent to  $\frac{\tan10^{\circ}}{\tan20^{\circ}\cdot \tan40^{\circ}}=\tan30^{\circ}$  or else

$$\tan20^{\circ}\cdot \tan30^{\circ}\cdot \tan40^{\circ}=\tan10^{\circ} \tag{1}$$

               Let's consider the equation

$$ \tan x\cdot \tan\frac{3x}{2}\cdot \tan 2x=\tan \frac{x}{2}\tag{2}$$

(1) shows that  $x=20^{\circ}$  is a solution for (2).

The problem of completely solving this equation remains open for now.

A forgotten trigonometric equation // Egy elfeledett trigonometrikus egyenlet

 We will solve the following trigonometric equation here :

$$\tan(30^{\circ}-x)\cdot \tan(30^{\circ}+x)=\frac{\tan x}{\sqrt{3}} \tag{1}$$

I first mentioned it in the Post here. There we also showed that the equation  (1)  is equivalent to the equation

$$t^3-3\sqrt{3}\cdot t^2-3\cdot t+\sqrt{3}=0,\;\;t=\tan x \tag{2}$$

ANSWER CiP

$$x_k^{\circ}=20^{\circ}+k\cdot 60^{\circ},\;\;k\in\mathbb{Z} \tag{3}$$

                    Solution CiP

              The calculation has already been done, but we are presenting it for convenience.

$(1)\overset{t=\tan x}{\Leftrightarrow} t^3-3\sqrt{3}\cdot t^2-3\cdot t+\sqrt{3}=0\Leftrightarrow \sqrt{3}=\frac{3t-t^3}{1-3t^2}\Leftrightarrow$

$\Leftrightarrow \sqrt{3}=\frac{3\cdot \tan x-\tan^3x}{1-3\cdot \tan^2x}\Leftrightarrow \tan 60^{\circ}=\tan 3x\Leftrightarrow 3x=60^{\circ}+k\cdot 180^{\circ},\;\;k\in \mathbb{Z}$

From here we get the answer  (3)

$\blacksquare$

          Remark CiP   It all started with solving the first equation from No. 6 on Page here. Now, after we have solved the equation  (1) (see also the Post from September 3, 2025), we can write the identities :

$$\frac{\tan10^{\circ}\cdot \tan50^{\circ}}{\tan20^{\circ}}=\frac{1}{\sqrt{3}}\tag{4}$$

$$\frac{\tan10^{\circ}\cdot \tan70^{\circ}}{\tan40^{\circ}}=\frac{1}{\sqrt{3}}\tag{5}$$

$$\frac{\tan50^{\circ}\cdot \tan70^{\circ}}{\tan80^{\circ}}=\frac{1}{\sqrt{3}}\tag{6}$$

          But  (4), considering that  $\frac{1}{\tan20^{\circ}}=\cot20^{\circ}$, is exactly the first identity from No. 6. Seeking to express all quantities as tangents of angles not exceeding  $45^{\circ}$, we further have  $\frac{\tan10^{\circ}\cdot \tan50^{\circ}}{\tan20^{\circ}}=\frac{\tan10^{\circ} \cdot \color{Red}{\cot40^{\circ}}}{\tan20^{\circ}}=\frac{\tan10^{\circ}\cdot \color{Red}{\frac{1}{\tan40^{\circ}}}}{\tan20^{\circ}}=\frac{\tan10^{\circ}}{\tan20^{\circ}\cdot \tan40^{\circ}}=\frac{1}{\sqrt{3}}$

which is the second entity from the aforementioned No. 6.

     Doing the same thing with  (5) : 

$\frac{1}{\sqrt{3}}=\frac{\tan10^{\circ}\cdot \color{Red}{\cot20^{\circ}}}{\tan40^{\circ}}=\frac{\tan10^{\circ}\cdot \color{Red}{\frac{1}{\tan20^{\circ}}}}{\tan40^{\circ}}=\frac{\tan10^{\circ}}{\tan20^{\circ}\cdot \tan{40^{\circ}}}$

and with (6) :

$\frac{1}{\sqrt{3}}=\frac{\color{Red}{\cot40^{\circ}\cdot \cot20^{\circ}}}{\color{Red}{\cot10^{\circ}}}=\frac{\frac{1}{\tan40^{\circ}}\cdot \frac{1}{\tan20^{\circ}}}{\frac{1}{\tan10^{\circ}}}=\frac{\tan10^{\circ}}{\tan20^{\circ}\cdot \tan40^{\circ}}$

So the second of the identities in No. 6 is the most important.

              Let's demonstrate it directly then. We will see that the pattern is the same.

$\frac{\tan10^{\circ}}{\tan20^{\circ}\cdot \tan40^{\circ}}=\frac{\frac{\sin10^{\circ}}{\cos10^{\circ}}}{\frac{\sin20^{\circ}}{\cos20^{\circ}}\cdot \frac{\sin40^{\circ}}{\cos40^{\circ}}}=\frac{8\cdot \sin10^{\circ}\cdot \cos20^{\circ}\cdot \cos40^{\circ}}{8\cdot \cos10^{\circ}\cdot \sin20^{\circ}\cdot \sin40^{\circ}}\tag{7}$

The numerator in (7) is  $4\cdot \frac{\sin20^{\circ}}{\cos10^{\circ}}\cdot \cos20^{\circ}\cdot \cos40^{\circ}=\frac{1}{\cos10^{\circ}}\cdot 2\cdot \sin40^{\circ}\cdot \cos40^{\circ}=\frac{1}{\cos10^{\circ}}\cdot \sin80^{\circ}=1$

The denominator in (7) is  $4\cdot \cos10^{\circ}\cdot(\cos20^{\circ}-\cos60^{\circ})=4\cos10^{\circ}\cdot \cos20^{\circ}-4\cos10^{\circ}\cdot \frac{1}{2}=$

$=2(\cos10^{\circ}+\cos30^{\circ})-2\cos10^{\circ}=2\cos10^{\circ}+2\cdot \frac{\sqrt{3}}{2}-2\cos10^{\circ}=\sqrt{3}$

So the final value in (7) is  $\frac{1}{\sqrt{3}}$.

<end Rem>

The Expressions for Roots of Cubic Polynomial // تعبيرات جذور كثيرة الحدود التكعيبية، مرة أخرى بمساعدة اللاهوت

                We have discussed this issue before. We will discuss the similarities and differences with the current situation at the end.

          Here we consider the problem : 

                   Let  $\tau$  be one of the roots of the equation

$t^3-3\sqrt{3}\cdot t^2-3\cdot t+\sqrt{3}=0 \tag{1}$

                   The other two roots are

$\tau_2=\frac{\sqrt{3}}{2}\cdot \tau^2-5\cdot \tau+\frac{\sqrt{3}}{2}\;,\;\;\tau_3=-\frac{\sqrt{3}}{2}\cdot \tau^2+4\cdot \tau+\frac{5\sqrt{3}}{2} \tag{2}$ 

SOLUTION  CiP

               From the fact that  $\tau$  checks  (1)  we obtain some useful formulas in future calculations

$\tau^3=3\sqrt{3}\cdot \tau^2+3\cdot \tau -\sqrt{3} \tag{3}$

$\tau^4=30\cdot \tau^2+8\sqrt{3} \cdot \tau-9 \tag{4}$

$\frac{1}{\tau}=-\frac{\sqrt{3}}{3}\cdot \tau^2+3\cdot \tau+\sqrt{3} \tag{5}$

          Indeed, $\tau^3-3\sqrt{3}\cdot \tau^2-3\cdot \tau +\sqrt{3}=0 \Rightarrow \tau^3=3\sqrt{3} \cdot \tau^2+3\cdot \tau-\sqrt{3}$,  i.e. (3). Then  $\tau^4=\tau \cdot \tau^3\overset{(3)}{=}$

$=\tau (3\sqrt{3}\cdot \tau^2+3\cdot \tau-\sqrt{3})=3\sqrt{3}\cdot \tau^3+3\cdot \tau^2-\sqrt{3}\cdot \tau\overset{(3)}{=}3\sqrt{3}(3\sqrt{3}\cdot \tau^2+3\cdot \tau-\sqrt{3})+3\cdot \tau^2-\sqrt{3}\cdot \tau=30\cdot \tau^2+8\sqrt{3}\cdot \tau-9$ i.e. (4). Finally, $\sqrt{3}=-\tau^3+3\sqrt{3}\cdot \tau^2+3\cdot \tau=\tau \cdot (-\tau^2+3\sqrt{3}\cdot \tau+3)\Rightarrow\frac{1}{\tau}=\frac{-\tau^2+3\sqrt{3}\cdot \tau+3}{\sqrt{3}}=-\frac{1}{\sqrt{3}}\cdot \tau^2+3\cdot \tau+\sqrt{3}$, i.e. (5).

          From the first formula of Vieta  $\tau_1+\tau_2+\tau_3=3\sqrt{3}$  we get

$$\tau_2+\tau_3=3\sqrt{3}-\tau \tag{6}$$

and from the third  $\tau_1\cdot \tau_2\cdot \tau_3=-\sqrt{3}$  we deduce

$$\tau_2\cdot \tau_3=-\frac{\sqrt{3}}{\tau} \tag{7}$$

(6) and (7) show that  $\tau_{2,3}$  are the roots of the quadratic equation

$t^2-(3\sqrt{3}-\tau) \cdot t-\frac{\sqrt{3}}{\tau}=0 \tag{8}$

          Let's calculate its discriminant:  $\Delta_2=(3\sqrt{3}-\tau)^2+\frac{4\sqrt{3}}{\tau}=$

$\overset{\color{Red}{!!!}}{=}27-6\sqrt{3}\cdot \tau+\tau^2+\frac{4\sqrt{3}\cdot \tau}{\color{Red}{\tau^2}}=\frac{\tau^4-6\sqrt{3}\cdot \tau^3+27\cdot \tau^2+4\sqrt{3}\cdot \tau}{\color{Red}{\tau^2}}=$

$\overset{(4)}{\underset{(3)}{=}}\frac{(30\cdot \tau^2+8\sqrt{3}\cdot \tau-9)-6\sqrt{3}(3\sqrt{3}\cdot \tau^2+3\cdot \tau-\sqrt{3})+27 \tau^2+4\sqrt{3}\cdot \tau}{\color{Red}{\tau^2}}=\frac{3\cdot \tau^2-6\sqrt{3}\cdot \tau+9}{\color{Red}{\tau^2}}=\frac{(\sqrt{3}\cdot \tau-3)^{\color{Red}2}}{\color{Red}{\tau^2}}$

<  Hallelujah, I got a perfect square  !!!  > 

Let's do a little more calculation to make writing easier :  $\pm\sqrt{\Delta_2}=\frac{\sqrt{3}\cdot \tau-3}{\tau}=$

$=(\sqrt{3}\cdot \tau-3)\cdot \frac{1}{\tau}\overset{(5)}{=}(\sqrt{3}\cdot \tau-3)\left (-\frac{\sqrt{3}}{3}\cdot \tau^2+3\cdot \tau+\sqrt{3}\right )=$

$=-\tau^3+\sqrt{3}\cdot \tau^2+3\sqrt{3}\cdot \tau^2-9\cdot \tau+3\cdot \tau-3\sqrt{3}=$

$\overset{(3)}{=}-(3\sqrt{3}\cdot \tau^2+3\cdot \tau-\sqrt{3})+4\sqrt{3}\cdot \tau^2-6\cdot \tau-3\sqrt{3}=\color{Blue}{\sqrt{3}\cdot \tau^2-9\cdot \tau-2\sqrt{3}}$

 !!! Remember this expression !!!

With these, the equation  (8)  has the roots

$\tau_{2,3}=\frac{3\sqrt{3}-\tau \pm \sqrt{\Delta_2}}{2}=\frac{3\sqrt{3}-\tau \pm (\sqrt{3}\cdot \tau^2-9\cdot \tau-2\sqrt{3})}{2} \tag{9}$

from where we obtain the expressions  (2).

$\blacksquare$

               REMARKS CiP

                    1.  In the calculation of  $\Delta_2$  that follows immediately after the equation  (8), I made a trick so that at least the denominator would be a perfect square. Then I was lucky enough !!  Compare with the Post  << Not only God does not help, but also Allah does not "put you in trouble">>.

If in (7) we had used (5), obtaining  $\tau_2\cdot \tau_3=\tau^2-3\sqrt{3}\cdot \tau-3$, then for the equation (8) we would have had  $\Delta_2=(3\sqrt{3}-\tau)^2-4(\tau^2-3\sqrt{3}\cdot \tau-3)=39+6\sqrt{3}\cdot \tau -3\cdot \tau^2.$  Seeing this expression of  $\Delta_2$  we wondered, as we did in the Post How? can we recognize a perfect square??, "how the hell" can we notice - compare with (9) - that

$$\color{Blue}{39+6\sqrt{3}\cdot \tau-3\cdot \tau^2=(\sqrt{3}\cdot \tau^2-9\cdot \tau-2\sqrt{3})^2}$$

Problem NOT solved yet.

                    2.  The theoretical part about when and how such a statement is possible was exposed in  the  Post "The Rational Expressions for Roots of Cubic Polynomial". There are some differences, but only apparent.

          First of all, that the polynomial  $f=X^3-3\sqrt{3}\cdot X^2-3\cdot X+\sqrt{3}$  is not in  $\mathbb{Q}[X]$  but in  $\mathbb{Q}(\sqrt{3})[X].$

          Then we need to make sure that the polynomial $f$  is irreducible. Maybe this is not that important, and I don't have an immediate justification at hand. 

          Third, its discriminant must be a perfect square (see Proposition in the Post). From  $f$  we obtain the depressed cubic by substitution  $t=u+\sqrt{3}$, obtainig the polynomial  $u^3-12\cdot u-8\sqrt{3}$. Its discriminant, the same as for  $f$, is  $-4p^3-27q^2=4\cdot12^3-27\cdot(64\cdot3)=1\;728$  so equal to $3\cdot 576=(24\sqrt{3})^2$, hence perfect square in  $\mathbb{Q}(\sqrt{3})$.

<end REM's>

Phương trình đại số có nghiệm lượng giác //Algebraic Equation with Trigonometric Solution

 We are showing here what we have left from a post from yesterday. I was helped for this in the AOPS Forum.

               The roots of the equation

$$t^3-3\sqrt{3}\cdot t^2-3\cdot t+\sqrt{3}=0 \tag{E}  $$

             are  $\tan20^{\circ},\;\;\tan80^{\circ},\;\;\tan140^{\circ}$

Adapted Solution by CiP

$(E)\Leftrightarrow \sqrt{3}\cdot (1-3t^2)=3t-t^3 \Leftrightarrow \sqrt{3}=\frac{3t-t^3}{1-3t^2}\;\underset{t=\tan x}{=}\frac{3\cdot \tan x-\tan^3x}{1-3\cdot \tan^2x}=\tan (3x)$

so, taking the equality of the extreme terms,  $3\cdot x=60^{\circ}+k\cdot 180^{\circ},\;\;k\in \mathbb{Z}$, hence  $x_k=20^{\circ}+k\cdot 60^{\circ}$. Returning to the variable  $t$,  we have the roots

$t_1=\tan( 20^{\circ}+3k\cdot 60^{\circ})=\tan (20^{\circ}+k\cdot 180^{\circ})=\tan20^{\circ},$

$t_2=\tan(20^{\circ}+(3k+1)\cdot 60^{\circ})=\tan(80^{\circ}+k\cdot 180^{\circ})=\tan80^{\circ}$

$t_3=\tan(20^{\circ}+(3k+2)\cdot 60^{\circ})=\tan(140^{\circ}+k\cdot 180^{\circ})=\tan140^{\circ}$.

$\blacksquare$

          Remark Cip  Vieta's formulas for the roots of the equation (E) lead to the equalities

$\tan20^{\circ}-\tan40^{\circ}+\tan80^{\circ}=3\sqrt{3}$

$\tan20^{\circ}\cdot \tan 40^{\circ}+\tan 40^{\circ}\cdot \tan 80^{\circ}-\tan 20^{\circ}\cdot \tan 80^{\circ}=3$

$\tan20^{\circ}\cdot \tan 40^{\circ}\cdot \tan80^{\circ}=\sqrt{3}$

<end Rem>

A trigonometric identity without variables

 We will prove the trigonometric identity

$$\tan10^{\circ}\cdot \tan50^{\circ}\cdot \cot20^{\circ}=\frac{1}{\sqrt{3}} \tag{1}$$

The problem appears at Solutions in GMB 3/2005, number O:1063, page 132. You can find more about this magazine here and here.

Official Solution

      Using that we have

 $\sin50^{\circ}=\cos(90^{\circ}-50^{\circ})=\cos40^{\circ},\;\;\cos10^{\circ}=\sin(90^{\circ}-10^{\circ})=\sin80^{\circ}$

and  $\cos50^{\circ}=\sin(90^{\circ}-50^{\circ})=\sin40^{\circ}$  we obtain

$\tan10^{\circ}\cdot \tan50^{\circ}\cdot \cot20^{\circ}=\frac{\sin10^{\circ}}{\cos10^{\circ}} \cdot \frac{\sin50^{\circ}}{\cos50^{\circ}} \cdot \frac{\cos 20^{\circ}}{\sin20^{\circ}}=\frac{\sin10^{\circ}\cdot \cos40^{\circ}\cdot \cos20^{\circ}}{\sin80^{\circ}\cdot \sin40^{\circ}\cdot \sin20^{\circ} } \tag{2}$

          Up in (2) we have, from  $\sin20^{\circ}=2\cdot \sin10^{\circ}\cdot \cos10^{\circ}$  if we replace  $2\cdot \sin10^{\circ}$ :

$8\cdot \sin10^{\circ}\cdot \cos20^{\circ}\cdot \cos40^{\circ}=\frac{4\cdot \sin20^{\circ}\cdot \cos20^{\circ}\cos40^{\circ}}{ \cos10^{\circ}}=\frac{2\cdot \sin40^{\circ}\cdot \cos40^{\circ}}{\cos10^{\circ}}=\frac{\sin80^{\circ}}{\cos10^{\circ}}=1.$

          Down in (2) we have, using  $2\sin u\sin v=\cos(v-u)-\cos(v+u)$, replacing  $\sin80^{\circ}=\cos10^{\circ}$, and using  $2\cos u \cos v=\cos(v-u)+\cos(v+u)$

$8\cdot \sin20^{\circ}\cdot \sin40^{\circ}\cdot \sin80^{\circ}=4(\cos20^{\circ}-\cos60^{\circ})\cdot \cos10^{\circ}=$

$=4\cdot \cos20^{\circ}\cdot \cos10^{\circ}-4\cdot \frac{1}{2}\cdot \cos10^{\circ}=2(\cos10^{\circ}+\cos30^{\circ})-2\cdot \cos10^{\circ}=$

$=2\cdot \cos10^{\circ}+2\cdot \frac{\sqrt{3}}{2}-2\cdot \cos10^{\circ}=\sqrt{3}$

These values ​​entered in  (2) give the answer.

$\blacksquare$

               Just as "After the battle, everyone is a general", we will also provide a solution.

                    Solution CiP

                Lemma CiP      $t_1=\tan20^{\circ}$  is  one of the roots of the equation

$$t^3-3\sqrt{3}\cdot t^2-3\cdot t+\sqrt{3}=0 \tag{3}$$

               Proof of Lemma

          In the triple angle formula  $\tan 3\theta=\frac{3\tan \theta-\tan^3 \theta}{1-3\tan^2 \theta}$  we substitute

$\theta=20^{\circ},\;\;t=\tan \theta$  and because  $\tan3\theta=\tan 60^{\circ}=\sqrt{3}$, we have

$\frac{3t-t^3}{1-3t^2}=\sqrt{3}\Leftrightarrow\;3t-t^3=\sqrt{3}-3\sqrt{3}\cdot t^2\Leftrightarrow\;t^3-3\sqrt{3}\cdot t^2-3\cdot t+\sqrt{3}=0$

that is (3).

qed Lemma$\square$

          To prove (1), we successively transcribe it (1)$\;\Leftrightarrow$

$\Leftrightarrow\;\frac{\tan(30^{\circ}-20^{\circ})\cdot \tan(30^{\circ}+20^{\circ})}{\tan20^{\circ}}=\frac{1}{\sqrt{3}}\Leftrightarrow\;\tan(30^{\circ}-20^{\circ})\cdot \tan(30^{\circ}+20^{\circ})=\frac{\tan20^{\circ}}{\sqrt{3}}$

and we have the equivalent problem: 

              Verify that the equation 

$$\tan(30^{\circ}-x)\cdot \tan (30^{\circ}+x)=\frac{\tan x}{\sqrt{3}} \tag{4}$$

               is satisfied by  $x=20^{\circ}$.

The left side of  (4)  is written :

$$\frac{\sin(30^{\circ}-x) \cdot \sin(30^{\circ}+x)}{\cos(30^{\circ}-x)\cdot \cos(30^{\circ}+x)}\;\;\overset{2\sin u\sin v=\cos((v-u)-\cos(v+u)}{\underset{2\cos u\cos v=\cos(v-u)+\cos(v+u)}{=}}$$

$=\frac{\cos2x-\cos60^{\circ}}{\cos2x+\cos60^{\circ}}\;\;\underset{\tan x=t}{=}\;\;\frac{\frac{1-t^2}{1+t^2}-\frac{1}{2}}{\frac{1-t^2}{1+t^2}+\frac{1}{2}}=\frac{2-2t^2-1-t^2}{2-2t^2+1+t^2}=\frac{1-3t^2}{3-t^2}$

so we need to show that the equation  $\frac{1-3t^2}{3-t^2}=\frac{t}{\sqrt{3}}$  is verified by  $t=\tan20^{\circ}$. But the equation with the unknown  $t$ is written equivalently

$\sqrt{3}-3\sqrt{3}\cdot t^2=3\cdot t-t^3\;\Leftrightarrow\;t^3-3\sqrt{3}\cdot t^2-3\cdot t+\sqrt{3}=0$

which the Lemma shows has the solution  $t=\tan20^{\circ}$.

$\blacksquare$

          Remark CiP  By this we have NOT completely solved the equation  (4) nor what are the other two roots of the equation  (3).

<end Rem>