Solve these equations simultaneously (trig)

Question

Solve for $ x,y: $

\begin{equation}\cos x -\cos(x+y) = 0 \end{equation}

\begin{equation}\cos y -\cos(x+y) = 0 \end{equation}

The answers are $(0, 0), (\frac{2\pi}{3}, \frac{2\pi}{3})$.

I get $(0, 0)$, but how do you get the latter?

Sorry if this is a bit basic - I don't remember trig well...

tfa

Answer 1

I should begin dividing first equation by cos(x), then second by cos(y), ....

Answer 2

Given

$$\cos x-\cos(x+y)=0\\ \cos y-\cos(x+y)=0$$

Subtracting the second equation from the first yields

$$\cos x=\cos y$$

Recall that $\cos x=\pm\sqrt{1-\sin^2x}$ and $\cos y=\pm\sqrt{1-\sin^2x}$.
This now becomes $\sqrt{1-\sin^2x}=\pm\sqrt{1-\sin^2y}$ yielding

$$\sin x=\pm\sin y$$

Recall that $\cos(x+y)=\cos x\cos y-\sin x\sin y$.
Thus, $\cos x-\cos(x+y)=0$ now becomes $\cos x-\cos^2x\pm\sin^2x=0$.
Solving $\cos x-\cos^2x-\sin^2x=0$ yields $\cos x=1$.
Solving $\cos x-\cos^2x+\sin^2x=0$ yields $\cos x-\cos^2x+1-\cos^2x=0$ which is equivalent to

$$(2\cos x+1)(-\cos x+1)=0$$

This yields two possibilities: $\cos x=1$ or $\cos x=-\frac12$. Can you do the rest?

Answer 3

We have $\cos x=\cos(x+y)=\cos y$

$\cos x=\cos y\implies x=2m\pi\pm y$ where $m$ is any integer

Case $\#1:$

If $x=2m\pi-y,\cos y=\cos2m\pi=1\implies y=2r\pi\equiv0\pmod{2\pi}$ where $r$ is any integer

$x\equiv-y\pmod{2\pi}\equiv-0$

Case $\#2:$

If $x=2m\pi+y,\cos y=\cos(x+y)=\cos(2m\pi+y+y)$

$\cos y=\cos2y\implies y=2n\pi\pm2y$ where $n$ is any integer

$'+'\implies y=2n\pi+2y\iff y=-2n\pi\equiv0\pmod{2\pi}$

$'-'\implies y=2n\pi-2y\iff y=\dfrac{2n\pi}3$