1. Let $t=\pi / 4-x / 3$. Then the original inequality is equivalent to:

$$
\begin{aligned}
3 \cos 2 t+2 \cos t-5 \geqslant 0 \Longleftrightarrow 3\left(2 \cos ^{2} t-1\right)+2 \cos t-5 \geqslant 0 \Longleftrightarrow \\
\Longleftrightarrow 6(\cos t-1)\left(\cos t+\frac{4}{3}\right) \geqslant 0
\end{aligned}
$$

We obtain $\cos t \in(-\infty ;-4 / 3] \cup[1 ;+\infty)$. Considering the range of cosine values, $\cos t=1$ or $\pi / 4-x / 3=2 \pi k, k \in \mathbb{Z}$.
Answer: $x=\frac{3 \pi}{4}+6 \pi k, k \in \mathbb{Z}$.

Answer to variant: $3-2: x=-\frac{3 \pi}{4}+6 \pi k, k \in \mathbb{Z}$.