# Solve system of equations with $\sin$ and $\cos$

Solve system of equations

$\begin{cases} 3x^2 + \sin 2y - \cos y - 3 = 0 \\ x^3 - 3x - \sin y - \cos 2y + 3 = 0 \end{cases}$

I tried to use substitution $x = \cos t$ or sth, but I get literally nothing

• Subtract $i$ times the first equation from the second equation, $$(x-i)^3-e^{2iy}+ie^{iy}+3+2i=0$$ – Empy2 Sep 11 '14 at 13:55
• One way to get rid of the trig is to substitute $t=\tan y/2$, then $\cos y =(1-t^2)/(1+t^2)$ and $\sin y=2t/(1+t^2)$ – Empy2 Sep 11 '14 at 14:09

The equations are $x^2=A$ and $x^3-3x=B$ where $$A=\frac{3+\cos y - \sin 2y}{3},\quad B=\sin y +\cos 2y - 3.$$ Now note that $x^3-3x=x(x^2-3),$ so one can substitute the value from $x^2=A$ here, obtaining $x(A-3)=B,$ that is, $x=B/(A-3).$ Then putting this $x$ back into $x^2=A$ gives the relation, involving only $A,B,$ that $$B^2=A(A-3)^2.\tag{1}$$ Here $A,B$ are trig functions of $y$ admitting period $2\pi,$ and a root finder indicates there are four solutions for $y$ namely $y=0.066078,\ 0.523599,\ 2.61799,\ 3.06375.$ In degrees, the second and third appear to be 30 and 150 respectively, while the first and last seem about 3.785 and 175.54 degrees, not that "nice" as solutions. (I haven't checked analytically whether the 30 and 150 degree solutions are exactly correct.)

Anyway, for each of the four possible $y$ values, one can compute its corresponding $x$ from $x=B/(A-3)$

Added: I checked the $y$ being 30 and 150 degree solutions and they are exact, each leading to $x=1.$ That is, in radians, two of the four solutions are $(x,y)=(1,\pi/6),\ (1,5\pi/6).$ Also the two other solutions for $y$ appear to add up to $\pi$ and each of those leads to the same value of $x$ for the other two pairs $(x,y)$ of solutions, though these others are not particularly nice. Going with the equation $(1)$ and trying to say put all in terms of cosine gives a high degree polynomial to solve, I think degree 8 once one uses $\sin^2 y + \cos^2 y=1$ to eliminate the other trig function.

Edit: Correction-- the "other two $y$ solutions" (besides $\pi/6,5\pi/6$) in fact do not add to $\pi$ but to about $3.1298.$ Furthermore their corresponding $x$ values are about $1.1352$ from the smaller $y$ and $0.8481$ from the larger. I noticed this while trying (unsuccessfully) to show that $B^2-A(A-3)^2$ was symmetric around $\pi/2$; it wasn't.

Invoking the double-angle formulas $\sin 2 y = 2 \sin y \cos y$ and $\cos 2 y = 1 - 2 \sin^2 y$, and abbreviating "$\sin y$" and "$\cos y$" by "$s$" and "$c$", we can manipulate the equations into these forms \begin{align} c \;( 2 s - 1 ) &= - 3 ( x^2 - 1 ) &= (x-1)\;P \tag{1}\\ s \;( 2 s - 1 ) &= -(x - 1 )^2 ( x + 2 ) &= (x-1)\;Q \tag{2} \end{align} where $P := -3(x+1)$ and $Q := -(x-1)(x+2)$. From here, we see a family of solutions that make each side of $(1)$ and $(2)$ vanish simultaneously:

$x=1$, and all $y$ such that $\sin y = \frac12$ (in particular, $y = \frac\pi6$ and $y = \frac{5\pi}{6}$)

We also see that there are no other solutions with either $x=1$ or $\sin y = \frac12$. Moreover, there are no other solutions with either $P=0$ or $Q=0$: if $x=-1$, then $(2)$ gives an impossible relation, $c(2s-1)=-4$; likewise, if $x=-2$, then $(1)$ gives $c(2s-1) = -9$. Consequently, we can write $$\frac{c}{P} = \frac{x-1}{2s-1} = \frac{s}{Q}$$ which implies $$c = \pm\frac{P}{R}\qquad s = \pm\frac{Q}{R}$$ where $R := \sqrt{P^2+Q^2}$ and the "$\pm$"s match.

Substituting into $(1)$ (or $(2)$), then, $$2 Q \mp R = ( x - 1 ) R^2 \quad\to\quad \left(\;2 Q - ( x - 1 )R^2\;\right)^2 = R^2$$ Rewriting $Q$ and $R^2 = P^2 + Q^2$ in terms of $x$, we have $$x^{10}+ 2 x^9 + 9 x^8 + 28 x^7 + 38 x^6 + 48 x^5 + 73 x^4 - 118 x^3 - 345 x^2 - 48 x + 276 = 0$$ for which Mathematica finds two real roots

$$x_1 \approx 1.1352 \qquad x_2 \approx 0.848153$$

The corresponding values of $Q/P$ give the tangents of the associated $y$s: $$\tan y_1 = \frac{-0.423884}{-6.4056} = \phantom{-}0.0661739$$ $$\tan y_2 = \frac{\phantom{-}0.432483}{-5.54446} = -0.0780028$$

One can check that

Only first-quadrant solutions $y_1$ and second-quadrant solutions $y_2$ actually satisfy the original equations.

Here's a graph of one period of the situation:

Equation $(1)$ defines the blue squiggles (right and middle); equation $(2)$ defines the red squiggle (left) and ovals. The line $x=1$ appears in gold.

Solving the first equation for $x$ $$3x^2+\sin(2y)-\cos(y)-3=0$$ $$x=\pm\sqrt{\frac{3+\cos(y)-\sin(2y)}{3}}$$ Substituting this into the second equation $$\left(\pm\sqrt{\frac{3+\cos(y)-\sin(2y)}{3}}\right)^3\mp3\sqrt{\frac{3+\cos(y)-\sin(2y)}{3}}-\sin(y)-\cos(2y)+3=0$$ Simplifying radicals $$\pm\frac{1}{3^{3/2}}(3+\cos(y)-\sin(2y))^{3/2}\mp\frac{3}{\sqrt{3}}(3+\cos(y)-\sin(2y))^{1/2}-\sin(y)-\cos(2y)+3=0$$ Expanding $\sin$'s, factoring the first two terms & ignoring $\pm,\mp$ signs to just take positive roots as this is already messy enough $$\frac{1}{\sqrt{3}}(3+\cos(y)-2\sin(y)\cos(y))^{1/2}\left(-2+\frac{1}{3}\cos(y)-\frac{2}{3}\sin(y)\cos(y)\right)-\sin(y)-\cos^2(y)+\sin^2(y)+3=0$$ Substituting $a=\sin(y)$, $b=\cos(y)$ $$\frac{1}{\sqrt{3}}(3+b-2ab)^{1/2}\left(\frac{1}{3}b-\frac{2}{3}ab-2\right)-a-b^2+a^2+3=0$$

We arrive at this messy (but solvable) equation. I can't think of an efficient way to solve it but solutions for the variables $a$ & $b$ can be computed with Wolfram Alpha at the following link

$\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ Solutions $\quad$ (wait a minute for it to load)

If you substitute the original forms of $a$ & $b$ in terms of $y$ each solution gives an equation of the form $\cos(y)=f(\sin(y))$

Sorry i couldn't come up with anything neater nor more general but the important thing is that a solution can be found. Also, I may be being ignorant & not noticing an easier solution. If there are any apparent errors in my work, please let me know. Thanks for the interesting question.

• We also have the constraint $a^2 + b^2 = 1$, which might simplify your final equation, although one has to be careful about signs. – Bungo Sep 10 '14 at 21:47
• @Bungo Yup. I can't think of a way to work it into what I've got so far. I'm still not entirely confident in what I've done yet. – Jam Sep 10 '14 at 21:51

Following on from coffeemath equation (1), and with the substitution $t=\tan(y/2)$, WolframAlpha gives the polynomial $$(t^2-4t+1)(5t^{10}-130t^9+66t^8-640t^7+128t^6-1044t^5+178t^4-656t^3+51t^2-122t+4)=0$$ The quadratic corresponds to $y=\pi/6$ and $y=5\pi/6$.

• A forbidding tenth degree polynomial left over! Does Wolfram have any success finding its (real) zeros? – coffeemath Sep 11 '14 at 14:48
• There is a pair of reals, given as decimals only; and eight complex roots scattered near the imaginary axis. – Empy2 Sep 11 '14 at 14:53