Solving $e^{ix}=i$

$$e^{ix}=i=e^{\frac{iπ}{2}}$$ $$\Longleftrightarrow e^{ix-\frac{iπ}{2}}=1$$ $$\Longleftrightarrow e^{i(x-\frac{π}{2})}=1$$ $$\Longleftrightarrow x-\frac{π}{2}=2nπ$$ $$\Longleftrightarrow x=2nπ+\frac{π}{2}$$

Hope it helps:)

You got the period wrong. The period of $sin(x)$ and $cos(x)$ is $2\pi$.

If $\cos(x) = 0$ and $\sin(x)=1$, therefore $$ x = \frac{\pi}{2} \text{ mod }2\pi $$ Thus $$ x= \frac{\pi}{2} + 2k\pi $$ where $k \in \mathbb{Z}$.