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I have the following trigonometric equation of $3$ variables:

$$f(\theta,\lambda,\phi)=3 \cos (\theta ) \cos (\lambda ) \cos (\phi )-(\cos (\theta )+3) \sin (\lambda ) \sin (\phi )$$

$$-\sin (\theta ) \cos (\lambda )+\sin (\theta ) \cos (\phi )+3 \cos (\theta )+\cos (\lambda ) \cos (\phi )-7$$

I want to prove that the solution set to the equation $f(\theta,\lambda,\phi)=0$ is

$$S_f=\{(2\pi k,\lambda+2\pi m,-\lambda+2\pi n):k,m,n\in\mathbb{Z},\lambda\in\mathbb{R}\}$$

Graphically, it is easy to see that this is the case but everything I have tried so far has failed to prove the conjecture. Perhaps my best attempt was extrapolating this equation into an unwieldy polynomial of $3$ variables

$$P(x,y,z)=9 x^4 y^2+16 x^4-18 x^3 y^2-192 x^3+9 x^2 y^4+z^4 \left(\left(x^2-1\right) y^2+1\right) \left(16 \left(x^2-1\right) y^2+(3 x+5)^2\right)-64 x^2 y^2+z^2 \left((x-1) (x+1) \left(9 x^2+30 x+41\right) y^4+2 x (x (48 (x-5) x+187)+234) y^2+x (x (9 (x-2) x-64)-78)-74 y^2+151\right)+736 x^2+2 y z^3 \left(9 x^4+12 x^3-20 x^2+(x-1) (x+1) (x (31 x-78)-33) y^2-96 x-33\right)+2 y z \left(31 x^4-270 x^3+560 x^2+(x (x (3 x (3 x+4)-20)-96)-33) y^2-210 x-111\right)+30 x y^4-78 x y^2-960 x+25 y^4+151 y^2+400$$

over the domain $(x,y,z)\in [-1,1]^3$. If I can prove that the solution set to $P(x,y,z)=0$ over this domain is

$$S_P=\{(1,y,y):y\in[-1,1]\}$$

then the original conjecture would be solved. The motivation behind this is actually proving a certain type of quantum error detection encoding exists. It's a little difficult to explain (although if anyone wants details I am more than happy to provide them) but suffice to say that after a lot of work I managed to whittle my existence proof down to the conjecture above.

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  • $\begingroup$ There's no way the solution set is what you claim it is. Why can't you have $(2\pi k,\lambda+2\pi, -\lambda)$? $\endgroup$ Jun 27, 2021 at 1:26
  • $\begingroup$ Fair enough, originally I was only considering the solutions over the space $[0,2\pi)^3$. When I expanded to $\mathbb{R}^3$ I forgot to update correctly $\endgroup$
    – QC_QAOA
    Jun 27, 2021 at 1:46
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    $\begingroup$ it's cool that your question comes from "proving a certain type of quantum error detection encoding exists"!! $\endgroup$ Jun 27, 2021 at 1:50
  • $\begingroup$ "Graphically, it is easy to see...": how can you display a function of $3$ variables ?? $\endgroup$
    – user65203
    Jun 28, 2021 at 16:37
  • $\begingroup$ I make a plot of $\lambda$ and $\phi$ and then I vary $\theta$ over $[0,2\pi]$ $\endgroup$
    – QC_QAOA
    Jun 28, 2021 at 16:39

4 Answers 4

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You can get a more tractable, hand-bashable quadratic equation if you do the famous half tangent substitution: $$\tan\frac\theta 2 = x,\,\, \tan\frac\phi 2 = y,\,\,\tan\frac\lambda 2 = z.$$ Then for example, one has: $$\cos\theta = \dfrac{1-x^2}{1+x^2}\text{ and } \sin\theta = \dfrac{2x}{1+x^2}.$$ This saves us the trouble of repeated squaring and once you multiply everything out, you will get a quadratic in $x$ whose discriminant is: $$\Delta = -16(y+z)^2(21y^2z^2+13y^2+13z^2+16yz+21)<0$$ unless $y+z = 0.$

The whole thing took me about 20 mins of hand checking, but once you know the plan of attack with the right substitution, it's nothing but guaranteed to be completed.

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We prove $f(\theta,\lambda,\phi) \le 0$ for each $\theta,\lambda,\phi$ and that equality holds exactly when you conjecture it does.

Write $$f(\theta,\lambda,\phi) = \cos(\theta)a+\sin(\theta)b+\cos(\lambda)\cos(\phi)-3\sin(\lambda)\sin(\phi)-7$$ for $a = 3\cos(\lambda)\cos(\phi)-\sin(\lambda)\sin(\phi)+3$ and $b = \cos(\phi)-\cos(\lambda)$. To maximize $f$, we of course to choose $\theta$ so that $\cos(\theta)a \ge 0$ and $\sin(\theta)b \ge 0$. So, for ease, let's just pretend $a,b \ge 0$ so that $\cos(\theta),\sin(\theta) \ge 0$. To maximize $xa+\sqrt{1-x^2}b$ for $x \in [0,1]$, one takes $x = \frac{a}{\sqrt{a^2+b^2}}$ (easy to prove by basic calculus), yielding a maximum of $\sqrt{a^2+b^2}$; and note that any other $x$ yields a strictly smaller value. Therefore, we wish to show $$\sqrt{(3\cos(\lambda)\cos(\phi)-\sin(\lambda)\sin(\phi)+3)^2+(\cos(\phi)-\cos(\lambda))^2}+\cos(\lambda)\cos(\phi)-3\sin(\lambda)\sin(\phi)-7 \le 0$$ with equality if and only if $\lambda = -\phi+2\pi m$ for some $m \in \mathbb{Z}$ (since then $\cos(\theta)$ will be $1$). For ease, let $x = \cos(\lambda)\cos(\phi)$ and $y = \sin(\lambda)\sin(\phi)$. We can rewrite the above as $$\sqrt{(3x-y+3)^2+(\cos(\phi)-\cos(\lambda))^2}+x-3y-7 \le 0.$$ We will show $$\sqrt{(3x-y+3)^2+(\cos(\phi)-\cos(\lambda))^2+(\sin(\phi)+\sin(\lambda))^2}+x-3y-7 \le 0$$ with equality if and only if $\lambda = -\phi+2\pi m$, which clearly suffices. The point of introducing the $(\sin(\phi)+\sin(\lambda))^2$ is that $$(\cos(\phi)-\cos(\lambda))^2+(\sin(\phi)+\sin(\lambda))^2 = 2-2x+2y.$$ So, we just wish to show that $$\sqrt{(3x-y+3)^2+2-2x+2y}+x-3y-7 \le 0$$ for $\{(x,y) \in [-1,1]^2 : |x-y| \le 1\}$ with equality if and only if $x-y = 1$. Note that $|3x-y+3| = |2x+(x-y)+3| \le 6$ and $|2-2x+2y| = 2|1-(x-y)| \le 4$, so $\sqrt{(3x-y+3)^2+2-2x+2y} \le \sqrt{40}$. Therefore, we cannot have $y \ge 0$ and $x \le 0$.

Let's first start with $y \ge 0$. Then, as just explained, $x \ge 0$. We will show that $$\sqrt{(3x-y+3)^2+2-2x+2y}+x-7 \le 0$$ with equality if and only if $x=1,y=0$. Note the derivative of the term inside the square root with respect to $y$ is $2y+2-6(x+1)$, which is negative, so we choose $y=0$ to maximize. We then wish to show $$\sqrt{(3x+3)^2+2-2x} -7 \le 0$$ with equality if and only if $x = 1$. But this is equivalent to $(3x+3)^2+2-2x \le 49$, which is true for $x \le 1$ with equality at and only at $x=1$.

Now let's deal with the case $x \le 0$. As explained before, this means $y \le 0$. We show $$\sqrt{(3x-y+3)^2+2-2x+2y}+x-3y-7 \le 0$$ for any $x,y \in [-1,0]^2$ with equality if and only if $y=-1,x=0$. The derivative of the term inside the square root with respect to $y$ is, as before, $2y+2-6(x+1)$, which is negative, so we take $y=-1$ to maximize. We then obtain $$\sqrt{9x^2+22x+16}+x-4,$$ which is maximized at $x=0$, yielding the value $0$. We're done.

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    $\begingroup$ It's not immediately clear where the other cases such as $a>0>b$ follows identically because their domains are technically different. For example if $a < 0<b,$ then your function to maximize is $-xa + \sqrt{1-x^2}b$ But I do think it will just be similar length bashing. $\endgroup$
    – dezdichado
    Jun 27, 2021 at 3:25
  • $\begingroup$ @dezdichado It does follow essentially identically. If $a < 0 < b$, we choose $\theta$ so that $\cos(\theta) < 0$ and $\sin(\theta) > 0$; or, we can just replace $x$ with $-x$. $\endgroup$ Jun 27, 2021 at 3:36
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Use Lagrange multipliers and Grobner bases to find the minimum and maximum of the function.

This will show that $f(\theta,\lambda,\phi)= 0$ is the maximum when $S_f=\{(2\pi k,\lambda+2\pi m,-\lambda+2\pi n):k,m,n\in\mathbb{Z},\lambda\in\mathbb{R}\} \tag{1}$

Let $x_1 = \cos(\theta)$, $y_1 = \sin(\theta)$. A constraint is $x_1^2 + y_1^2 - 1 = 0$.

Similarly $x_2= \cos(\lambda)$, $y_2= \sin(\lambda)$, $x_3= \cos(\phi)$, $y_3= \sin(\phi)$.

$$F(x_1,y_1,x_2,y_2,x_3,y_3) = 3x_{1}x_{2}x_{3} -\left(x_{1}+3\right)y_{2}y_{3}-x_{2}y_{1}+x_{3}y_{1}+3x_{1} +x_{2}x_{3}-7 \tag{2}$$

The constraints are:

$$g_1 = x_1^2 + y_1^2 - 1 \,, g_2 = x_2^2 + y_2^2 - 1 \,, g_3 = x_3^2 + y_3^2 - 1 \tag{3}$$

The Lagrangian is: $$\mathcal{L} = F - L_1g_1 - L_2g_2 -L_3g_3 \tag{4}$$

Solve: $$\nabla \mathcal{L} = 0 \tag{5}$$

Maxima:

G(x1,y1,x2,y2,x3,y3,L1,L2,L3) :=  3*x1*x2*x3 - (x1+3)*y2*y3
 -y1*x2 + y1*x3 +3*x1 + x2*x3 - 7 - L1*(x1^2 + y1^2 - 1) 
- L2*(x2^2 + y2^2 - 1) -L3*(x3^2 + y3^2 - 1);

J : jacobian([G(x1,y1,x2,y2,x3,y3,L1,L2,L3)],[x1,y1,x2,y2,x3,y3,L1,L2,L3]);
J : transpose(J);

$$\nabla \mathcal{L} = \begin{pmatrix}-\mathit{y_2}\, \mathit{y_3}+3 \mathit{x_2}\, \mathit{x_3}-2 \mathit{L_1}\, \mathit{x_1}+3\\ -2 \mathit{L_1}\, \mathit{y_1}+\mathit{x_3}-\mathit{x_2}\\ -\mathit{y_1}+3 \mathit{x_1}\, \mathit{x_3}+\mathit{x_3}-2 \mathit{L_2}\, \mathit{x_2}\\ -\left( \mathit{x_1}+3\right) \, \mathit{y_3}-2 \mathit{L_2}\, \mathit{y_2}\\ \mathit{y_1}-2 \mathit{L_3}\, \mathit{x_3}+3 \mathit{x_1}\, \mathit{x_2}+\mathit{x_2}\\ -2 \mathit{L_3}\, \mathit{y_3}-\left( \mathit{x_1}+3\right) \, \mathit{y_2}\\ -{{\mathit{y_1}}^{2}}-{{\mathit{x_1}}^{2}}+1\\ -{{\mathit{y_2}}^{2}}-{{\mathit{x_2}}^{2}}+1\\ -{{\mathit{y_3}}^{2}}-{{\mathit{x_3}}^{2}}+1\end{pmatrix} \tag{6}$$

Find a Grobner basis:

Maxima:

load(grobner);
vars : [x1,y1,x2,y2,x3,y3,L1,L2,L3];
eqns : transpose(J)[1];
gb : poly_reduced_grobner(eqns,reverse(vars));
transpose(gb);

The Grobner basis:

$$\begin{pmatrix} -y_{1}^2- x_{1}^2+1\cr -y_{2}^2-x_{2}^2+1\cr x_{1}-x_{1}^3\cr x_{1}^2y_{2}- y_{2}\cr 3y_{3}^2+4x_{1}x_{2}y_{1}+3x_{2}^2-3\cr x_{1} y_{2}y_{3}-4x_{1}^2x_{2}y_{1}+2x_{2}y_{1}-3x_{1} x_{2}^2-3x_{1}+2L_{1}\cr 3x_{1}y_{2}y_{3}+9y_{2}y_{3}- 6x_{1}^2x_{2}y_{1}-2x_{1}x_{2}y_{1}+3x_{2}y_{1}-9 x_{1}x_{2}^2-3x_{2}^2-6x_{1}^2+6L_{2}+6\cr -24x_{2}y_{3} -3x_{1}y_{1}y_{2}-7y_{1}y_{2}-24x_{1}x_{2}y_{2}\cr 3 x_{1}y_{2}y_{3}+9y_{2}y_{3}-12x_{1}^2x_{2}y_{1}-2 x_{1}x_{2}y_{1}+3x_{2}y_{1}-9x_{1}x_{2}^2-3x_{2}^2-6 x_{1}^2+6L_{3}+6\cr y_{1}y_{3}+x_{1}y_{1}y_{2}\cr x_{1}^2 y_{3}-y_{3}\cr 2x_{1}y_{1}-3x_{3}+6x_{1}^2x_{2}-3x_{2} \cr -3x_{2}^2y_{1}-6x_{1}^2y_{1}-2x_{1}y_{1}+3y_{1} \cr -x_{1}^2x_{2}^2+x_{2}^2+x_{1}^2-1\cr \end{pmatrix} \tag{7}$$

The third equation gives $x_1(x_1^2-1) = 0$. So the maximum and minimum occur at $x_1 = [-1,0,1]\tag{8}$

Consequently $x_1y_1 = 0$ due to constraint $g_1$ equation $(3)$.

The fifth equation gives $3y_{3}^2+4x_{1}x_{2}y_{1}+3x_{2}^2-3 = 0$. Substituting $x_1y_1 = 0$ gives $y_{3}^2+x_{2}^2-1 = 0$. This is a unity constraint use ($g_2,g_3$) which give $x_2^2 = x_3^2 \tag{9}$ $y_2^2 = y_3^2\tag{10}$

So we can add equations to the basis: $[x_1y_1,x_2^2 - x_3^2, y_2^2 - y_3^2]$.

Recalculate the basis:

Maxima:

eqns2 : append(eqns,[x1*y1,x2^2-x3^2,y2^2-y3^2]);   
gb2 : poly_reduced_grobner(eqns2,reverse(vars));
gb2 : transpose(gb2);

$$\begin{pmatrix} -y_{1}^2- x_{1}^2+1\cr -y_{2}^2-x_{2}^2+1\cr x_{1}\,y_{1}\cr x_{1}-x_{1}^3\cr x_{3}-2\,x_{1}^2\,x_{2}+x_{2}\cr y_{1}\,y_{2}\cr y_{1}-x_{2}^2\, y_{1}\cr x_{1}^2\,x_{2}^2-x_{2}^2-x_{1}^2+1\cr x_{1}^2\,y_{2}-y_{2} \cr y_{3}^2+x_{2}^2-1\cr -x_{1}\,y_{2}\,y_{3}-3\,y_{2}\,y_{3}-x_{2} \,y_{1}+3\,x_{1}\,x_{2}^2+x_{2}^2+2\,x_{1}^2-2\,L_{3}-2\cr -x_{1}\, y_{2}\,y_{3}-2\,x_{2}\,y_{1}+3\,x_{1}\,x_{2}^2+3\,x_{1}-2\,L_{1}\cr x_{2}\,y_{3}+x_{1}\,x_{2}\,y_{2}\cr -x_{1}\,y_{2}\,y_{3}-3\,y_{2}\, y_{3}-x_{2}\,y_{1}+3\,x_{1}\,x_{2}^2+x_{2}^2+2\,x_{1}^2-2\,L_{2}-2 \cr y_{1}\,y_{3}\cr x_{1}^2\,y_{3}-y_{3}\cr \end{pmatrix} \tag{11}$$

Equations $6$ and $15$ give: $y_{1}\,y_{2} = 0 \tag{12}$ $y_{1}\,y_{3} = 0 \tag{13}$ These may be useful later but I cannot see any more useful reductions.

At this point brute force is an option. $(x_1,y_1):(-1,0),(0,1),(0,-1),(1,0)$. $x_2 = x_3$, $x_2 = -x_3$. $y_2 = y_3$, $y_2 = -y_3$. Which gives $4\times 2 \times 2 = 16$ test cases. The number of cases will be reduced when $(x_1,y_1)$ are added to the Grobner basis and recalculated.

The function to evaluate:

Maxima:

F(x1,y1,x2,y2,x3,y3) :=  3*x1*x2*x3 - (x1+3)*y2*y3 -y1*x2 + y1*x3 +3*x1 + x2*x3 - 7;

$ ------------------------------------$

case $(x_1,y_1) = (-1,0)$.

There should be no solutions for this case.

Add $x_1$ and $y_1$ to the basis $(11)$.

Maxima:

eqns_minus1_0 : append(eqns2,[x1+1,y1]);    
gb_minus1_0 : poly_reduced_grobner(eqns_minus1_0,reverse(vars));
gb_minus1_0 : transpose(gb_minus1_0);

$$\begin{pmatrix} -y_{2}^2- x_{2}^2+1\cr x_{1}+1\cr y_{1}\cr x_{3}-x_{2}\cr y_{2}\,y_{3}+x_{2}^2 +L_{3}\cr x_{2}\,y_{3}-x_{2}\,y_{2}\cr y_{2}\,y_{3}+x_{2}^2+L_{2} \cr y_{3}^2+x_{2}^2-1\cr -y_{2}\,y_{3}+3\,x_{2}^2+2\,L_{1}+3\cr \end{pmatrix} \tag{14}$$

The fourth equation gives $x_2 = x_3$. So the only test cases required are $y_3 = y_2$ and $y_3 = -y_2$ (see equation $(10)$).

case $y_3 = y_2$

$F(-1,0,x2,y2,x2,y2) = -2\,y_{2}^2-2\,x_{2}^2-10 = -2\,(y_{2}^2+\,x_{2}^2)-10 = -12$

By $g_2$ $(3)$.

This is a local minima and not a solution to $F = 0$.

case $y_3 = -y_2$:

$F(-1,0,x2,y2,x2,-y2) = 2\,y_{2}^2-2\,x_{2}^2-10 = -4\,x_{2}^2-8$

This is an inverted parabola with a maximum of $-8$ at $x_2 = 0$.

There are no solutions to $F = 0$ for the case $(x_1,y_1) = (-1,0)$.

$ ------------------------------------$

case $(x_1,y_1) = (0,-1)$.

There should be no solutions for this case.

Add $x_1$ and $y_1$ to the basis $(11)$.

Maxima:

eqns_0_minus1 : append(eqns2,[x1,y1+1]);    
gb_0_minus1 : poly_reduced_grobner(eqns_0_minus1,reverse(vars));
gb_0_minus1 : transpose(gb_0_minus1);

$$\begin{pmatrix} x_{1}\cr y_{1}+1\cr x_{2}-2\,L_{3}-1\cr x_{3}+x_{2}\cr 1-x_{2}^2\cr y_{2}\cr L_{1}-x_{2}\cr y_{3}\cr -x_{2}+2\,L_{2}+1\cr \end{pmatrix} \tag{15}$$

From the fourth equation $x_3 = -x_2$.

From the sixth equation $y_2 = 0$.

From the eighth equation $y_3 = 0$.

$F(0,-1,x2,0,-x2,0) = -x_{2}^2+2\,x_{2}-7$

This is an inverted parabola with a minimum of $F = -6$ at $x_2 = 1$.

There is no solution to $F = 0$ for the case $(x_1,y_1) = (0,-1)$.

$ ------------------------------------$

case $(x_1,y_1) = (0,1)$.

There should be no solutions for this case.

Add $x_1$ and $y_1$ to the basis $(11)$.

Maxima:

eqns_0_1 : append(eqns2,[x1,y1-1]); 
gb_0_1 : poly_reduced_grobner(eqns_0_1,reverse(vars));
gb_0_1 : transpose(gb_0_1);

$$\begin{pmatrix} x_{1}\cr y_{1}-1\cr -x_{2}-2\,L_{3}-1\cr x_{3}+x_{2}\cr x_{2}^2-1\cr y_{2} \cr x_{2}+L_{1}\cr y_{3}\cr x_{2}+2\,L_{2}+1\cr \end{pmatrix} \tag{16}$$

From the fourth equation $x_3 = -x_2$.

From the sixth equation $y_2 = 0$.

From the eighth equation $y_3 = 0$.

$F(0,1,x2,0,-x2,0) = -x_{2}^2-2\,x_{2}-7$

This is an inverted parabola with a maximum of $F = -6$ at $x_2 = -1$.

There is no solution to $F = 0$ for the case $(x_1,y_1) = (0,1)$.

$ ------------------------------------$

case $(x_1,y_1) = (1,0)$.

There should be solutions for this case.

Add $x_1$ and $y_1$ to the basis $(11)$.

Maxima:

eqns_1_0 : append(eqns2,[x1-1,y1]); 
gb_1_0 : poly_reduced_grobner(eqns_1_0,reverse(vars));
gb_1_0 : transpose(gb_1_0);

$$\begin{pmatrix} -y_{2}^2- x_{2}^2+1\cr x_{1}-1\cr y_{1}\cr x_{3}-x_{2}\cr 2\,y_{2}\,y_{3}-2\, x_{2}^2+L_{3}\cr x_{2}\,y_{3}+x_{2}\,y_{2}\cr 2\,y_{2}\,y_{3}-2\, x_{2}^2+L_{2}\cr y_{3}^2+x_{2}^2-1\cr -y_{2}\,y_{3}+3\,x_{2}^2-2\, L_{1}+3\cr \end{pmatrix} \tag{17}$$

From the fourth equation $x_3 = x_2$. So the only test cases required are $y_3 = y_2$ and $y_3 = -y_2$.

case $y_3 = y_2$.

$F(1,0,x2,y2,x2,y2) = -4\,y_{2}^2+4\,x_{2}^2-4 = 8\,x_{2}^2-8$

The maximum occurs when $x_2 = \pm 1$ then $F = 0$. Note $y_2 = 0$ by $g_2$ equation $(3)$. Note $0 = -0$. Note $y_3 = y_2 = 0 = -y_3$. The only solution in this case still obeys $(x_1,y_1,x_2,y_2,x_3,y_3) = (1,0,x_2,y_2,x_2,-y_2)$.

case $y_3 = -y_2$.

$F(1,0,x2,y2,x2,-y2) = 4\,y_{2}^2+4\,x_{2}^2-4 = 0$.

By $g_2$ equation $(3)$.

$F = 0$ whenever $(x_1,y_1,x_2,y_2,x_3,y_3) = (1,0,x_2,y_2,x_2,-y_2)$.

$ ------------------------------------$

The only solutions to $F = 0$ occur at $(x_1,y_1,x_2,y_2,x_3,y_3) = (1,0,x_2,y_2,x_2,-y_2)$.

This translates to $x_1 = \cos(\theta) = 1$ so $\theta = 2\pi k$.

$x_2 = x_3$ translates to $\cos(\lambda) = \cos(\phi)$.

$y_2 = - y_3$ translates to $\sin(\lambda) = -\sin(\phi)$.

From the identities $\cos(\alpha) = \cos(-\alpha)$ and $\sin(\alpha) = -\sin(-\alpha)$ with periodicity of $2\pi$.

These conditions are represented by $S_f=\{(2\pi k,\lambda+2\pi m,-\lambda+2\pi n):k,m,n\in\mathbb{Z},\lambda\in\mathbb{R}\}$.

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Let $\;x=\tan\dfrac\theta2,\;y=\dfrac{\lambda+\varphi}2,\;z=\dfrac{\lambda-\varphi}2,\;$ then $$g(x,y,z) = (1-x^2)(2\cos2y+\cos 2z+3)+4x\sin y\sin z\\ +(1+x^2)(-\cos2y+2\cos2z-7)\\ = (-3\cos2y+\cos2z-10)x^2+4x\sin y\sin z+\cos2y+3\cos2z-4\\ = (-6\cos^2y-2\sin^2z-6)x^2+4x\sin y\sin z - 2\sin^2y-6\sin^2z=0,$$ $$-6(1+\cos^2y)x^2-2(x\sin z-\sin y)^2-6\sin^2z=0.\tag1$$ From $(1)$ should $$x=0,\quad y=\pi j,\quad z=\pi l,\tag2$$ and finally $$\color{green}{\mathbf{\theta=2\pi k,\quad \lambda=\pi m,\quad \varphi=2\pi n-\lambda,\quad k,m,n\in\mathbb Z.}}$$

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