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MATU • Trigonometria
MATU_TREC_090
Propio
Enunciado
Resolver la ecuación:
$$ \tan \left( \frac{\pi}{4} + x \right) + \tan \left( \frac{\pi}{4} - x \right) = 4 $$
$$ \tan \left( \frac{\pi}{4} + x \right) + \tan \left( \frac{\pi}{4} - x \right) = 4 $$
Solución Paso a Paso
1. Aplicación de identidades de suma de ángulos:
Recordamos que $\tan(\frac{\pi}{4}) = 1$. Usamos $\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$:
$$ \frac{1 + \tan x}{1 - \tan x} + \frac{1 - \tan x}{1 + \tan x} = 4 $$
2. Operación de fracciones:
Sea $u = \tan x$:
$$ \frac{1+u}{1-u} + \frac{1-u}{1+u} = 4 $$
$$ \frac{(1+u)^2 + (1-u)^2}{(1-u)(1+u)} = 4 $$
$$ \frac{(1 + 2u + u^2) + (1 - 2u + u^2)}{1 - u^2} = 4 $$
$$ \frac{2 + 2u^2}{1 - u^2} = 4 $$
3. Despeje de la variable:
$$ 2 + 2u^2 = 4(1 - u^2) \Rightarrow 2 + 2u^2 = 4 - 4u^2 $$
$$ 6u^2 = 2 \Rightarrow u^2 = \frac{2}{6} = \frac{1}{3} $$
$$ u = \pm \sqrt{\frac{1}{3}} = \pm \frac{1}{\sqrt{3}} $$
4. Cálculo de los ángulos:
Si $\tan x = \frac{1}{\sqrt{3}}$:
$$ x = \frac{\pi}{6} $$
Si $\tan x = -\frac{1}{\sqrt{3}}$:
$$ x = \frac{5\pi}{6} $$
5. Resultado final:
$$ \boxed{x = \frac{\pi}{6}} $$
Recordamos que $\tan(\frac{\pi}{4}) = 1$. Usamos $\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$:
$$ \frac{1 + \tan x}{1 - \tan x} + \frac{1 - \tan x}{1 + \tan x} = 4 $$
2. Operación de fracciones:
Sea $u = \tan x$:
$$ \frac{1+u}{1-u} + \frac{1-u}{1+u} = 4 $$
$$ \frac{(1+u)^2 + (1-u)^2}{(1-u)(1+u)} = 4 $$
$$ \frac{(1 + 2u + u^2) + (1 - 2u + u^2)}{1 - u^2} = 4 $$
$$ \frac{2 + 2u^2}{1 - u^2} = 4 $$
3. Despeje de la variable:
$$ 2 + 2u^2 = 4(1 - u^2) \Rightarrow 2 + 2u^2 = 4 - 4u^2 $$
$$ 6u^2 = 2 \Rightarrow u^2 = \frac{2}{6} = \frac{1}{3} $$
$$ u = \pm \sqrt{\frac{1}{3}} = \pm \frac{1}{\sqrt{3}} $$
4. Cálculo de los ángulos:
Si $\tan x = \frac{1}{\sqrt{3}}$:
$$ x = \frac{\pi}{6} $$
Si $\tan x = -\frac{1}{\sqrt{3}}$:
$$ x = \frac{5\pi}{6} $$
5. Resultado final:
$$ \boxed{x = \frac{\pi}{6}} $$