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3cot(x)+1=2+4cot(x),0<= x<= 360

Solución

3 cot (x) + 1 = 2 + 4 cot (x), 0^{\circ} \leq x \leq 36 0^{\circ}

x = 13 5^{\circ}, x = 31 5^{\circ}

radianes

x = \frac{3 π}{4}, x = \frac{7 π}{4}

Pasos de solución

3 cot (x) + 1 = 2 + 4 cot (x), 0^{\circ} \leq x \leq 36 0^{\circ}

Usando el método de sustitución

3 cot (x) + 1 = 2 + 4 cot (x)

Sea:

cot (x) = u

3 u + 1 = 2 + 4 u

3 u + 1 = 2 + 4 u : u = - 1

3 u + 1 = 2 + 4 u

Mover

1

al lado derecho

3 u + 1 = 2 + 4 u

Restar

1

de ambos lados

3 u + 1 - 1 = 2 + 4 u - 1

Simplificar

3 u = 4 u + 1

3 u = 4 u + 1

Mover

4 u

al lado izquierdo

3 u = 4 u + 1

Restar

4 u

de ambos lados

3 u - 4 u = 4 u + 1 - 4 u

Simplificar

- u = 1

- u = 1

Dividir ambos lados entre

- 1

- u = 1

Dividir ambos lados entre

- 1

\frac{- u}{- 1} = \frac{1}{- 1}

Simplificar

u = - 1

u = - 1

Sustituir en la ecuación

u = cot (x)

cot (x) = - 1

cot (x) = - 1

cot (x) = - 1, 0 \leq x \leq 36 0^{\circ} : x = 13 5^{\circ}, x = 31 5^{\circ}

cot (x) = - 1, 0 \leq x \leq 36 0^{\circ}

Soluciones generales para

cot (x) = - 1

cot (x)

tabla de valores periódicos con

18 0^{\circ} n

intervalos:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

x = 13 5^{\circ} + 18 0^{\circ} n

x = 13 5^{\circ} + 18 0^{\circ} n

Soluciones para el rango

0 \leq x \leq 36 0^{\circ}

x = 13 5^{\circ}, x = 31 5^{\circ}

Combinar toda las soluciones

x = 13 5^{\circ}, x = 31 5^{\circ}

sin (x) = \frac{48.35}{50.14}

2 tan (\frac{π x}{2}) = sec (\frac{π x}{2})

1.47 = (\frac{sin ( \frac{x + 60}{2} )}{sin ( 3 0 ^{\circ} )})

sin^{2} (x) + 1 = 3 sin (x) - 1

2 tan (θ) = sec^{2} (θ)