3

The question is to solve

$$\cos\left(2x-\frac{\pi}{3}\right)=\cos(x)$$

I originally approached this using the addition formulae but the mark scheme showed a way by first replacing $x$ on the right with $2\pi-x$ and I understand this is due to the $\cos $ graph, however I don't understand where all the solutions came from, would really appreciate help understanding

Answers: $\frac{\pi}{3}, \frac{7\pi}{3}$ and $\frac{13\pi}{3}$.

Thanks

Michael Hardy
  • 1
Radhika
  • 31

3 Answers3

1

Note that because of the $2\pi$-periodicity and evenness of the cosine function, we have $\cos(x)=\cos(\pm x+2k\pi)$ for all integer $k$.

If $\cos(2x-\pi/3)=\cos(x)$, then $2x-\pi/3=\pm x+2k\pi$, whereupon we find

$$x=\frac{\pi/3+2k\pi}{2\pm 1}$$

Mark Viola
  • 179,405
0

$$\cos { \left( 2x-\frac { \pi }{ 3 } \right) =\cos { x } } $$ $$2x-\frac { \pi }{ 3 } =x+2\pi n,\quad \quad \Rightarrow x=\frac { \pi }{ 3 } +2\pi n,\quad $$ $$2x-\frac { \pi }{ 3 } =-x+2\pi n,n=0,\pm 1,\pm 2,..\quad \Rightarrow x=\frac { \pi }{ 9 } +\frac { 2\pi n }{ 3 } ,n=0,\pm 1\quad ,\pm 2,..$$

haqnatural
  • 21,578
0

There are other solutions. To see this, it's simpler to use congruences: the basic equations in trigonometry are

  • $\cos a=\cos b\iff a\equiv \pm b\mod 2\pi$,
  • $\sin a=\sin b\iff a\equiv\begin{cases}b\\\pi-b\end{cases}\mod 2\pi,$
  • $\tan a=\tan b\iff a\equiv b\mod \pi$.

Here you obtain $$2x-\frac\pi3\equiv\pm x\mod2\pi\iff\begin{cases}x\equiv \dfrac\pi3\mod 2\pi\\[1ex]3x\equiv \dfrac\pi3\mod2\pi\end{cases}\iff\begin{cases}x\equiv \dfrac\pi3\mod 2\pi\\[1ex]x\equiv \dfrac\pi9\mod\dfrac{2\pi}3\end{cases}$$

Bernard
  • 175,478