1
$\begingroup$

We have to solve for $x$ and $y$:

$$\left \{ \begin{align*} \cos (2x+3y) &= \frac{1}{2} \\ \cos (3x+2y) &= \frac{\sqrt{3}}{2} \end{align*} \right.$$

I got $$\left \{ \begin{align*} 2x+3y &= 2n\pi \pm\frac{\pi}{3} \\ 3x+2y &= 2m\pi \pm\frac{\pi}{6} \end{align*} \right.$$

The final answer in the book is $x=\dfrac{1}{5} \left[ (6m-4n)\pi \pm \dfrac{\pi}{2} \mp \dfrac{2\pi}{3} \right]$

Why is it $\mp \dfrac{2\pi}{3}$?

If for some value we had taken

$$2x+3y=2n\pi -\frac{\pi}{3}$$

and

$$3x+2y=2m\pi +\frac{\pi}{6}$$

The value of $x$ would be

$$\frac{1}{5}[(6m-4n)\pi+\frac{\pi}{2}+\frac{2\pi}{3}]$$

which is not included in the general solution.

  • 0
    is this an equation system? – 2017-01-10
  • 1
    **In this case**, the notation $\pm a\mp b$ refers to the set $$\{a+b,a-b,-a+b,-a-b\}$$ hence it is equivalent to the notation $\pm a\pm b$ interpreted in the same way (two signs hence four possibilities). Do not ask me to justify its usage, which I think is faulty because of its ambiguities. Sometimes, $\pm a\mp b$ refers to the set $$\{a-b,-a+b\}$$ and this (the sign chosen once and for all hence two possibilities instead of four) is probably what you had in mind. – 2017-01-10

2 Answers 2

1

$1)$ You have:

$$2x+3y = 2n\pi \pm\frac{\pi}{3}$$

Multiply by $-2$ and get:

$$-4x-6y = -4n\pi \mp\frac{2\pi}{3} \quad (I)$$

$2)$ You have: $$ 3x+2y= 2m\pi \pm\frac{\pi}{6}$$

Multiply by $3$ and get: $$9x+6y= 6m\pi \pm \frac{\pi}{2}\quad (II)$$

$(I)+(II)$:

$$5x=(6m-4n)\pi \mp\frac{2\pi}{3}\pm \frac{\pi}{2}$$

PS: $\mp\frac{2\pi}{3}$ means that you got opposite signals when multiplied by a negative number, in that case it was $-2$.

0

To get rid of $y$, the first equation is multiplied by 2 and the second by 3 giving $\pm 2\pi/3\pm\pi/2$.

To get rid of $x$,the first equation is multiplied by 3 and the second by 2 giving $\pm \pi\pm\pi/3$.