<s>[INST] Solve trigonometric equation $ 3 \cos x + 2\sin x=1 $ Solve trigonometric equation: $$ 3 \cos x + 2\sin x=1 $$
I tried to substitue $\cos x = \dfrac{1-t^2}{1+t^2}, \sin x = \dfrac{2t}{1+t^2}$. Yet with no results.
 [/INST] 
$$ 3 \cos x + 2\sin x=1 $$

$y:=\tan\big(\frac x 2\big)$ then $\sin(x)=\frac{2y}{y^2+1}$ and $\cos(x)=\frac{1-y^2}{y^2+1}$
$$-1+\frac{3}{y^2+1}+\frac{4y}{y^2+1}-\frac{3y^2}{y^2+1}=0$$
$$\frac{2y^2-2y-1}{y^2+1}=0$$
$$2y^2-2y-1=0$$
$$y^2-y=\frac 1 2$$
Add $\frac 1 4$ to both sides:
$$y^2-y+\frac 1 4=\frac 3 4$$
$$\bigg(y-\frac 1 2 \bigg)^2=\frac 3 4$$
$y=\frac 1 2+\frac{\sqrt 3}{2}$ or $y=\frac{\sqrt 3}{2}+\frac 1 2$
$y=\tan\big(\frac{\pi}{2}\big)$
$$\boxed{\color{blue}{x=2\arctan\big(\frac 1 2\pm\frac{\sqrt 3}{2}+2 n \pi\big)\;\;\;\;, n\in \mathbb{Z} }}$$
</s>