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.
 A: Try this:
$$3 \cos x + 2\sin x=\sqrt{13}\left(\frac{3}{\sqrt{13}}\cos x+\frac{2}{\sqrt{13}}\sin x\right)\\
=\sqrt{13}\sin(\arcsin\frac{3}{\sqrt{13}}+x)=1$$
You can try to solve it from there.
A: If you have a sum $S=a\cos x+b \sin x$ 
define $r$ by $r^2=a^2+b^2$ 
and $\theta$ by $r\sin \theta =a$ and $r\cos \theta =b$ so that $\tan \theta =\cfrac ab$
Then $S=r\sin \theta \cos x+r\cos \theta \sin x=r\sin (x+\theta)$
A: Let $u = \cos x$ and $v = \sin x$. Thus, $3u + 2v = 1$ and $u^2 + v^2 = 1$. Isolating one of the variables , we have
$$
u^2 + \dfrac{(1 - 3u)^2}{4} = 1 \quad \Rightarrow \quad 13u^2 - 6u - 3 = 0 
$$
and so on...
A: 
$$ 3 \cos x + 2\sin x=1 $$

$y:=\tan\big(\frac x 2\big)$ 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 $y=\frac{\sqrt 3}{2}+\frac 1 2$
$y=\tan\big(\frac{\pi}{2}\big)$
$$\boxed{\color{blue}{x=2\arctan\big(\frac 1 2\pm\frac{\sqrt 3}{2}+2 n \pi\big)\;\;\;\;, n\in \mathbb{Z} }}$$
