Question

3 tg (2x + pi/4) + √3 = 0

Answer 1

$\mathtt{3tg(2x+\frac{\pi}{4})+\sqrt{3}=0;~tg(2x+\frac{\pi}{4})=-\frac{\sqrt{3}}{3}}$

$\mathtt{2x+\frac{\pi}{4}=-\frac{\pi}{6}+\pi n,n\in Z}$

следовательно, наш корень $\mathtt{x=\frac{12n\pi-5\pi}{24},n\in Z}$