# Math Help - LAST one - Promise! Trigonometric Ecuations.

1. ## LAST one - Promise! Trigonometric Ecuations.

I know that how to work with cos, sin and tan...

tanˆ2x + tanx = 0
tanx(tanx + 1) = 0

tanx = 0
x = undefined

tanx + 1 = 0
tanx = -1
x = -45

But, how do I solve:

(√3)csc(x) - 2 = 0

Thank you!

2. Hello, Polyxendi!

$\tanˆ2\!x + \tan x \:=\: 0$

$\tan x(\tan x + 1) \:=\: 0$

$\tan x \:=\: 0 \quad\Rightarrow\quad x \:=\:\text{unde{f}ined}$ . . . . no . . ${\color{red} x \:=\:\hdots\:\text{-}180^o,\:0^o,\:180^o,\:360^o,\:\hdots}$

$\tan x + 1 \:=\: 0\quad\Rightarrow\quad \tan x \:=\: \text{-}1\quad\Rightarrow\quad{\color{blue} x \:=\:\hdots \text{-}225^o,\:\text{-}45^o,\:135^o,\:315^o,\:\hdots}$

But how do I solve: . $\sqrt{3}\csc x - 2 \:=\: 0$
Same way . . .

We have: . $\sqrt{3}\csc x \:=\:2 \quad\Rightarrow\quad \csc x \:=\:\frac{2}{\sqrt{3}}$

You're expected to recognize that as: $x \,=\,60^o$

Therefore: . $x \;=\;\hdots\:\text{-}240^o,\:60^o,\:120^o,\:420^o,\:\hdots$

3. Originally Posted by Polyxendi
I know that how to work with cos, sin and tan...

tanˆ2x + tanx = 0
tanx(tanx + 1) = 0

were you given an interval to find solutions for x ?

tanx = 0
x = undefined no, that's not correct

tanx + 1 = 0
tanx = -1
x = -45 that is one solution, x = what other values?

But, how do I solve:

(√3)csc(x) - 2 = 0

$\textcolor{red}{\csc{x} = \frac{2}{\sqrt{3}}}$

$\textcolor{red}{\sin{x} = \frac{\sqrt{3}}{2}}$

solutions for the last equation are straight off the unit circle, which you have memorized, right ?

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