Đáp án: `x=-\frac{5π}{36}+\frac{kπ}{3} \ (k∈\mathbb{Z})`
Giải:
Đkxđ:
`cos(3x+\frac{π}{6}) \ne 0`
⇔ `3x \ne \frac{π}{2}+kπ`
⇔ `x \ne \frac{π}{6}+\frac{kπ}{3} \ (k∈\mathbb{Z})`
Ta có:
`tan(3x+\frac{π}{6})=-1`
⇔ `tan(3x+\frac{π}{6})=tan(-\frac{π}{4})`
⇔ `3x+\frac{π}{6}=-\frac{π}{4}+kπ`
⇔ `3x=-\frac{π}{6}-\frac{π}{4}+kπ`
⇔ `3x=-\frac{5π}{12}+kπ`
⇔ `x=-\frac{5π}{36}+\frac{kπ}{3} \ (k∈\mathbb{Z})` (thỏa)
