1. ## New trigo-system

Solve :
$\displaystyle sin(x)cos(y)=\frac{1}{4}$
$\displaystyle 3Tan(x)=Tan(y)$

2. Originally Posted by dhiab
Solve :
$\displaystyle sin(x)cos(y)=\frac{1}{4}$
$\displaystyle 3Tan(x)=Tan(y)$
Infinitely many solutions.

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

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

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

we have

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

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

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

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

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

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

Therefore

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

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

4. Hello, dhiab!

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

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

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

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

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

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

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

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

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