459. a) $\sin x=\frac{\sqrt{3}}{2}$; b) $\cos x=\frac{\sqrt{2}}{2}$; c) $\operatorname{tg} x=3$; d) $\operatorname{ctg} x=-1$.
Solution. a) According to formula (24), we get

$$
x=(-1)^{k} \arcsin \frac{\sqrt{3}}{2}+\pi k=(-1)^{k} \frac{\pi}{3}+\pi k, \quad k \in \mathbf{Z}
$$

b) By formula (25) we find

$$
x= \pm \arccos \frac{\sqrt{2}}{2}+2 \pi k= \pm \frac{\pi}{4}+2 \pi k, \quad k \in \mathbf{Z}
$$

c) In accordance with formula (26) we have

$$
x=\operatorname{arcctg} 3+\pi k, \quad k \in \mathbf{Z}
$$

d) Applying formula (27) and considering that $\operatorname{arcctg}(-1)=\pi-$ $-\operatorname{arcctg} 1=\pi-\frac{\pi}{4}=\frac{3 \pi}{4}$, we get

$$
x=\operatorname{arcctg}(-1)+\pi k=\frac{3 \pi}{4}+\pi k, \quad k \in \mathbf{Z}
$$