\( 123 \cos ^{2} x+\mathrm{R} 3 \sin x \cos x=0 \); find GS

We start with the equation \[ 123 \cos^2 x + \sqrt{3}\, \sin x \cos x = 0. \] Notice that both terms contain a factor of \(\cos x\). Factor it out: \[ \cos x \left(123 \cos x + \sqrt{3}\, \sin x\right) = 0. \] This gives us two cases: 1. \(\cos x = 0\) 2. \(123 \cos x + \sqrt{3}\, \sin x = 0\) --- **Case 1:** \(\cos x = 0\) The general solution for \(\cos x = 0\) is: \[ x = \frac{\pi}{2} + k\pi,\quad k\in\mathbb{Z}. \] --- **Case 2:** \(123 \cos x + \sqrt{3}\, \sin x = 0\) Assuming \(\cos x \neq 0\), divide the equation by \(\cos x\): \[ 123 + \sqrt{3}\, \tan x = 0. \] Solve for \(\tan x\): \[ \tan x = -\frac{123}{\sqrt{3}}. \] Simplify the fraction by multiplying the numerator and denominator by \(\sqrt{3}\): \[ \tan x = -\frac{123\sqrt{3}}{3} = -41\sqrt{3}. \] The general solution for \(\tan x = -41\sqrt{3}\) is: \[ x = \arctan(-41\sqrt{3}) + k\pi,\quad k\in\mathbb{Z}. \] Alternatively, using the odd function property of arctan, this may also be written as: \[ x = -\arctan(41\sqrt{3}) + k\pi,\quad k\in\mathbb{Z}. \] --- **Final General Solution:** \[ \boxed{ \begin{aligned} x &= \frac{\pi}{2} + k\pi,\quad k\in\mathbb{Z},\\[1mm] \text{or}\quad x &= \arctan(-41\sqrt{3}) + k\pi,\quad k\in\mathbb{Z}. \end{aligned} } \]