# Thread: solve for x algebraically

1. ## solve for x algebraically

solve x algebraically not graphically

3 cot (90 + x) = tan x sin x

Thanx

2. $\displaystyle 3\cot (90+x) = \tan x \sin x$

$\displaystyle 3\frac{\cos (90 + x)}{\sin (90 + x)} = \frac{\sin x}{\cos x} \sin x$

$\displaystyle 3\frac{-\sin x}{\cos x} = \frac{\sin^2 x}{\cos x}$

$\displaystyle 3\frac{-\sin x}{\not{\cos x}} = \frac{\sin^2 x}{\not{\cos x}}$

$\displaystyle -3\sin x = \sin^2 x$

.........$\displaystyle \sin x = 0$ works

We'll try to get another answer,

$\displaystyle -3\not{\sin x} = \sin^{\not 2} x$

$\displaystyle -3 = \sin x$ There's no solution here.

Then we only use $\displaystyle \sin x = 0$.

Hence, $\displaystyle x = k\pi$

3. Hello, Nico!

A slightly different approach . . .

$\displaystyle 3\cot(90 + x) \:=\: \tan x\sin x$

Since $\displaystyle \cot(x + 90) \:=\:-\tan x$, we have:

. . $\displaystyle -3\tan x \:=\:\tan x\sin x \quad\Rightarrow\quad \tan x\sin x + 3\tan x \:=\:0$

Factor: .$\displaystyle \tan x(\sin x + 3) \:=\:0$

But .$\displaystyle \sin x + 3 \:=\:0\quad\Rightarrow\quad \sin x \:=\:-3\;\text{ has no solutions}$

Therefore: .$\displaystyle \tan x \:=\:0\quad\Rightarrow\quad x \:=\:\pi n\:\text{ for any integer }n$