Question

Can anyone explain those marked steps in case (iii)

Answer 1

$\cos\dfrac{3x}2=\cos\left(\dfrac\pi2+\dfrac x2\right)$
$\implies \dfrac{3x}2=2n\pi+\dfrac\pi2+\dfrac x2$

follows from the fact that the cosine function is $2\pi$-periodic, which means $\cos x=\cos(2\pi+x)$. Roughly speaking, this is the same as saying that a point on a circle is the same as the point you get by completing a full revolution around the circle (i.e. add $2\pi$ to the original point's angle with respect to the horizontal axis).

If you make another complete revolution (so we're effectively adding $4\pi$) we get the same result: $\cos x=\cos(4\pi+x)$. This is true for any number of complete revolutions, so that this pattern holds for any even multiple of $\pi$ added to the argument. Therefore $\cos x=\cos(2n\pi+x)$ for any integer $n$.

Next, because $\cos(-x)=\cos x$, it follows that $\cos x=\cos(2n\pi-x)$ is also true for any integer $n$. So we have

$\implies \dfrac{3x}2=2n\pi\pm\left(\dfrac\pi2+\dfrac x2\right)$

The rest follows from considering either case and solving for $x$.