Solve the system of equations $$\begin{cases}163-400z\sin{x}&=0\\-135z+85\cos{x}+61&=0\end{cases}$$

What is the best way of going about this?

I rearranged the second equation for $z$ and then substituted it into the first one to find $$163-400\left(\frac{1}{135}(85\cos{x}+61)\right)\sin{x}=0$$ $$\implies 22005 - 34000\sin{x}\cos{x}-24400\sin{x}=0$$ $$\implies 4401-6800\sin{x}\cos{x}-4880\sin{x}=0$$ $$\implies 80\sin{x}(85\cos{x}-61)=4401$$ Now how do I solve this?

  • $\begingroup$ There will be probably not a nice solution. But it exists, because for $x=0$ the left hand side of the last equation is positive and for $x=\pi/2$ it is negative. $\endgroup$ – Peter Franek Feb 4 '16 at 18:41
  • $\begingroup$ we have $\sin(2x)=2\sin(x)\cos(x)$ $\endgroup$ – Dr. Sonnhard Graubner Feb 4 '16 at 18:42
  • $\begingroup$ @Dr.SonnhardGraubner Sure, but is it useful? (Interestingly, your post below does not even mention this identity.) $\endgroup$ – Did Feb 7 '16 at 8:29

Writing $s = \sin(x)$ and $c = \cos(x)$, the resultant of $4401 -6800 s c - 4800 s$ and $c^2 + s^2 - 1$ with respect to $c$ is $$ 46240000 s^4-22425600 s^2-42953760 s+19368801 $$ which is an irreducible quartic over the rationals. So $\sin(x)$ is a root of that polynomial. It does have a solution in terms of radicals, but this is not pleasant to look at. There are two real roots, approximately $ .3955840368$ and $.9976236060$.

EDIT: The corresponding values of $z$ according to your first equation are then approximately $1.030122457$ and $.4084706873$ respectively, and the values of $\cos(x)$ from your second are approximately $.9184297849$ and $-.0688994967$. The corresponding approximate values of $x$ are then $.4067036720 + 2 \pi n$ and $1.639750453 + 2 \pi n$ for integers $n$.

EDIT: The real roots of that quartic can be expressed as follows. Let $$ \eqalign{a_1 &= 134802424991908+7803\sqrt{176073951178517210885}\cr a_2 &= a_1^{2/3} + 18688 a_1^{1/3} + 1953171499\cr a_3 &= \sqrt{\sqrt{a_2}(37376 a_1^{1/3}-a_1^{2/3}-1953171499)+18255348 \sqrt{2 a_1}}\cr }$$ Then the values of $\sin(x)$ are $$ \dfrac{a_2^{3/4} \pm a_3}{680\; a_1^{1/6} a_2^{1/4}}$$


HINT: use that $$\sin(x)=2\,{\frac {\tan \left( x/2 \right) }{1+ \left( \tan \left( x/2 \right) \right) ^{2}}} $$ and $$\cos(x)={\frac {1- \left( \tan \left( x/2 \right) \right) ^{2}}{1+ \left( \tan \left( x/2 \right) \right) ^{2}}} $$

  • $\begingroup$ why that? these are known identities look at the internet. Why the $-1$? $\endgroup$ – Dr. Sonnhard Graubner Feb 4 '16 at 18:52
  • 2
    $\begingroup$ @MichaelChirico Those identities are true, but I fear they are not very useful in this case. $\endgroup$ – Robert Israel Feb 4 '16 at 18:54
  • $\begingroup$ they are useful also in this case $\endgroup$ – Dr. Sonnhard Graubner Feb 4 '16 at 18:55
  • 1
    $\begingroup$ Indeed -- these can transform the problem to a fourth-order polynomial in $\tan(\frac{x}2)$. Still won't give a nice neat solution, but I like the idea. $\endgroup$ – MichaelChirico Feb 4 '16 at 19:01

Robert Israel gave the rigorous solution of the problem solving the quartic equation in $s$.

Outside this approach, only numerical methods would lead to the solution; probably, Newton method could be the simplest.

Looking at the plot of function $$f(s)=46240000 s^4-22425600 s^2-42953760 s+19368801$$ $(-1 \leq s \leq 1)$, you can notice that there is one root close to $0.4$ and another close to $1.0$.

So, let us use Newton method with these starting guesses.

For the first root, the iterates will be $$x_1=0.395575435005952$$ $$x_2=0.395584036761436$$ $$x_3=0.395584036792970$$ which is the solution for fifteen significant figures.

For the second root, the iterates will be $$x_1=0.997638403524923$$ $$x_2=0.997623606632696$$ $$x_3=0.997623606053749$$ which is the solution for fifteen significant figures.


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