If I have $\cos(x)=-\cos(x+\alpha)$,
can I solve it by doing
$x=-(x+\alpha+2\pi)$ and $x=-(-(x+\alpha+2\pi))$?
It's probably a stupid question but I'm really confused.
If I have $\cos(x)=-\cos(x+\alpha)$,
can I solve it by doing
$x=-(x+\alpha+2\pi)$ and $x=-(-(x+\alpha+2\pi))$?
It's probably a stupid question but I'm really confused.
Hint:
$\cos(u+\pi)=-\cos u$
and
$\cos x=\cos A\implies x=2m\pi\pm A$
Where $m$ is any integer
HINT: $$\cos(x)+\cos(y)=2\cos\left(\frac{x-y}{2}\right)\cos\left(\frac{x+y}{2}\right)$$