Can someone explain this? $\sec(x/2) = \cos(x/2)$
I worked on this and got here...
(Let (x/2) = u)
$\cos u - \sec u = 0$
$\cos u(1 - \sec^2u) = 0$
$\cos u[ -1(-1 + \sec^2u)] = 0$
$\cos u(-\tan^2u) = 0$
So, the solutions would be:
$x = pi + 4\pi k, 3\pi + 4\pi k, 0 + 2\pi k$ but the problem is that the first two $(\pi + 4\pi k,3\pi + 4 \pi k)$end up making the original equation have an undefined term $(
\sec(x/2))$. Is this simply because I went out of terms of the original equation? If so does this mean that every time I go out of terms of the original I must check the answers? This is confusing me a lot because usually you don't have to check answers unless you square both sides. 

As $\displaystyle\cos^2\frac x2=1,\cos x=2\cos^2\frac x2-1=1=\cos0$
$\displaystyle\implies x=2n\pi\pm0$ where $n$ is any integer

Another way : $\displaystyle\cos^2\frac x2=\cos^2\frac A2$
$\displaystyle\implies \frac x2=m\pi\pm\frac A2\iff x=2m\pi\pm A$ where $m$ is any integer
Here $A=0$
