Ii
MATU • Trigonometria
MATU_TREC_075
2do Ex. II-2012
Enunciado
Halle las soluciones de la ecuación, entre $0^\circ$ y $360^\circ$:
$$ \frac{\cos^2 x + \tan x \operatorname{sen} x}{3 + \sec x} = \frac{1}{4} $$
$$ \frac{\cos^2 x + \tan x \operatorname{sen} x}{3 + \sec x} = \frac{1}{4} $$
Solución Paso a Paso
1. Simplificación del numerador:
Sustituimos $\tan x = \frac{\operatorname{sen} x}{\cos x}$:
$$ \cos^2 x + \frac{\operatorname{sen} x}{\cos x} \cdot \operatorname{sen} x = \frac{\cos^3 x + \operatorname{sen}^2 x}{\cos x} $$
2. Simplificación del denominador:
Sustituimos $\sec x = \frac{1}{\cos x}$:
$$ 3 + \frac{1}{\cos x} = \frac{3\cos x + 1}{\cos x} $$
3. Sustitución en la ecuación original:
$$ \frac{\frac{\cos^3 x + \operatorname{sen}^2 x}{\cos x}}{\frac{3\cos x + 1}{\cos x}} = \frac{1}{4} \Rightarrow \frac{\cos^3 x + \operatorname{sen}^2 x}{3\cos x + 1} = \frac{1}{4} $$
Usamos $\operatorname{sen}^2 x = 1 - \cos^2 x$:
$$ \frac{\cos^3 x - \cos^2 x + 1}{3\cos x + 1} = \frac{1}{4} $$
4. Resolución algebraica:
Multiplicamos en cruz:
$$ 4\cos^3 x - 4\cos^2 x + 4 = 3\cos x + 1 $$
$$ 4\cos^3 x - 4\cos^2 x - 3\cos x + 3 = 0 $$
Factorizamos por agrupación:
$$ 4\cos^2 x (\cos x - 1) - 3(\cos x - 1) = 0 $$
$$ (\cos x - 1)(4\cos^2 x - 3) = 0 $$
5. Hallar valores de $x$:
$$ \boxed{x = \{0^\circ; 30^\circ; 150^\circ; 210^\circ; 330^\circ; 360^\circ\}} $$
Sustituimos $\tan x = \frac{\operatorname{sen} x}{\cos x}$:
$$ \cos^2 x + \frac{\operatorname{sen} x}{\cos x} \cdot \operatorname{sen} x = \frac{\cos^3 x + \operatorname{sen}^2 x}{\cos x} $$
2. Simplificación del denominador:
Sustituimos $\sec x = \frac{1}{\cos x}$:
$$ 3 + \frac{1}{\cos x} = \frac{3\cos x + 1}{\cos x} $$
3. Sustitución en la ecuación original:
$$ \frac{\frac{\cos^3 x + \operatorname{sen}^2 x}{\cos x}}{\frac{3\cos x + 1}{\cos x}} = \frac{1}{4} \Rightarrow \frac{\cos^3 x + \operatorname{sen}^2 x}{3\cos x + 1} = \frac{1}{4} $$
Usamos $\operatorname{sen}^2 x = 1 - \cos^2 x$:
$$ \frac{\cos^3 x - \cos^2 x + 1}{3\cos x + 1} = \frac{1}{4} $$
4. Resolución algebraica:
Multiplicamos en cruz:
$$ 4\cos^3 x - 4\cos^2 x + 4 = 3\cos x + 1 $$
$$ 4\cos^3 x - 4\cos^2 x - 3\cos x + 3 = 0 $$
Factorizamos por agrupación:
$$ 4\cos^2 x (\cos x - 1) - 3(\cos x - 1) = 0 $$
$$ (\cos x - 1)(4\cos^2 x - 3) = 0 $$
5. Hallar valores de $x$:
- $\cos x = 1 \Rightarrow x = \{0^\circ, 360^\circ\}$
- $\cos^2 x = \frac{3}{4} \Rightarrow \cos x = \pm \frac{\sqrt{3}}{2}$
- $\cos x = \frac{\sqrt{3}}{2} \Rightarrow x = \{30^\circ, 330^\circ\}$
- $\cos x = -\frac{\sqrt{3}}{2} \Rightarrow x = \{150^\circ, 210^\circ\}$
$$ \boxed{x = \{0^\circ; 30^\circ; 150^\circ; 210^\circ; 330^\circ; 360^\circ\}} $$