Question

$\begin{array}{l}\mathsf{Obter~tg~x~~sabendo~que~~sin^2x-5\cdot sin~x\cdot cos~x~+cos^2x=3}\end{array}$

$\mathsf{gab:~~tg~x=-2~~ou~~tg~x=\dfrac{-1}{2}}$

*Trigonometria, relações fundamentais.

Answer 1

Olá Vinícius, boa noite!
 
Fiz assim:

$\\ \mathsf{\sin^2 x - 5 \cdot \sin x \cdot \cos x + \cos^2 x = 3} \\\\ \mathsf{\underbrace{\mathsf{\sin^2 x + \cos^2 x}}_{1} - 5 \cdot \sin x \cdot \cos x = 3} \\\\ \mathsf{- 5 \cdot \sin x \cdot \cos x = 3 - 1} \\\\ \mathsf{\sin x \cdot \cos x = - \frac{2}{5}} \\\\ \mathsf{\sin x = - \frac{2}{5 \cdot \cos x}}$
 
 Queremos encontrar $\mathsf{\tan x}$, isto é,

$\\ \mathsf{\tan x = \frac{\sin x}{\cos x}} \\\\\\ \mathsf{\tan x = - \frac{2}{5 \cdot \cos x} \div \cos x} \\\\\\ \mathsf{\tan x = - \frac{2}{5 \cdot \cos x} \times \frac{1}{\cos x}} \\\\\\ \mathsf{\tan x = - \frac{2}{5 \cdot \cos^2 x} \qquad \qquad (i)}$
 
 Recorrendo à relação $\mathsf{\sin^2 x + \cos^2 x = 1}$, temos que:

$\\ \mathsf{\sin^2 x + \cos^2 x = 1} \\\\ \mathsf{\left ( - \frac{2}{5 \cdot \cos x} \right )^2 + \cos^2 x = 1} \\\\\\ \mathsf{\frac{4}{25 \cdot \cos^2 x} + \cos^2 x = 1} \\\\\\ \mathsf{25 \cdot \cos^4 x - 25 \cdot \cos^2 x + 4 = 0}$ 

$\\ \mathsf{25 \cdot \cos^4 x - 20 \cdot \cos^2 x - 5 \cdot \cos^2 x + 4 = 0} \\\\ \mathsf{5 \cdot \cos^2 x \cdot \left ( 5 \cdot \cos^2 x - 4 \right ) - 1 \cdot \left ( 5 \cdot \cos^2 x - 4 \right ) = 0} \\\\ \mathsf{\left ( 5 \cdot \cos^2 x - 4 \right ) \cdot \left [ 5 \cdot \cos^2 x - 1 \right ] = 0}$
 
 Com efeito,

$\\ \mathsf{\left ( 5 \cdot \cos^2 x - 4 \right ) \cdot \left [ 5 \cdot \cos^2 x - 1 \right ] = 0} \\\\ \Downarrow \\\\ \begin{cases} \mathsf{5 \cdot \cos^2 x - 4 = 0 \Rightarrow \boxed{\mathsf{\cos^2 x = \frac{4}{5}}}} \\ \mathsf{5 \cdot \cos^2 x - 1 = 0 \Rightarrow \boxed{\mathsf{\cos^2 x = \frac{1}{5}}}} \end{cases}$
 
 Por fim, substituímos... em (i). Segue,

$\\ \bullet \qquad \mathsf{Quando \ \boxed{\mathsf{\cos^2 x = \frac{4}{5}}}:} \\\\\\ \mathsf{\tan x = - \frac{2}{5} \cdot \frac{1}{\cos^2 x}} \\\\\\ \mathsf{\tan x = - \frac{2}{5} \cdot \frac{1}{\frac{4}{5}}} \\\\\\ \mathsf{\tan x = - \frac{2}{5} \cdot \frac{5}{4}} \\\\\\ \boxed{\boxed{\mathsf{\tan x = - \frac{1}{2}}}}$
 
$\\ \bullet \bullet \qquad \mathsf{Quando \ \boxed{\mathsf{\cos^2 x = \frac{1}{5}}}:} \\\\\\ \mathsf{\tan x = - \frac{2}{5} \cdot \frac{1}{\cos^2 x}} \\\\\\ \mathsf{\tan x = - \frac{2}{5} \cdot \frac{1}{\frac{1}{5}}} \\\\\\ \mathsf{\tan x = - \frac{2}{5} \cdot \frac{5}{1}} \\\\\\ \boxed{\boxed{\mathsf{\tan x = - 2}}}$
 
 Boa questão!!