1. New trigo-system

Solve :
$
sin(x)cos(y)=\frac{1}{4}
$

$
3Tan(x)=Tan(y)
$

2. Originally Posted by dhiab
Solve :
$
sin(x)cos(y)=\frac{1}{4}
$

$
3Tan(x)=Tan(y)
$
Infinitely many solutions.

3. $\sin(x) \cos(y) = \frac{1}{4}$

$3\tan(x) = \tan(y)$ or

$3\sin(x)\cos(y) = \sin(y)\cos(x) = \frac{3}{4}$

we have

$\sin(y)\cos(x) = \frac{3}{4}$ and

$\sin(x) \cos(y) = \frac{1}{4}$

$\sin(x+y) = \sin(x) \cos(y) + sin(y)\cos(x) = 1$

$\sin(y-x) = sin(y)\cos(x) - \sin(x) \cos(y) = \frac{1}{2}$

$x+y = \frac{4n+1}{2} \pi$

$y-x = m\pi + (-1)^m \frac{\pi}{6}$

Therefore

$x = \frac{1}{2}(\frac{4n+1}{2} \pi - m\pi - (-1)^m \frac{\pi}{6} )$

$y = \frac{1}{2}(\frac{4n+1}{2} \pi + m\pi + (-1)^m \frac{\pi}{6} )$

4. Hello, dhiab!

I will assume that $x$ and $y$ are acute angles.
You can generalize them later.

Solve: . $\begin{array}{cccc}
\sin x\cos y &=& \frac{1}{4} & [1] \\ 3\tan x&=& \tan y & [2] \end{array}$

From [2], we have: . $3\frac{\sin x}{\cos x} \:=\:\frac{\sin y}{\cos y} \quad\Rightarrow\quad \cos x\sin y \:=\:3\sin x\cos x$

Substitute [1]: . $\cos x\sin y \:=\:3\left(\tfrac{1}{4}\right)$

. . $\begin{array}{cccccc}\text{We have:} & \cos x\sin y &=& \frac{3}{4} \\ \\[-3mm]
\text{Add [1]:} & \sin x\cos y &=& \frac{1}{4} \end{array}$

And we have: . $\sin x\cos y + \cos x \sin y \:=\:\frac{3}{4} + \frac{1}{4} \quad\Rightarrow\quad \sin(x+y) \:=\:1 \quad\Rightarrow\quad x + y \:=\:\frac{\pi}{2}$

. . $x$ and $y$ are complementary . . $\Rightarrow\quad \cos y \:=\:\sin x$

Substitute into [1]: . $\sin x\cdot\sin x \:=\:\frac{1}{4} \quad\Rightarrow\quad \sin x \:=\:\frac{1}{2}$

Therefore: . $x \:=\:\frac{\pi}{6},\;\;y \:=\:\frac{\pi}{3}$