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solvefor x,cot^2(x)= 1/3

Solución

resolver para

x, cot^{2} (x) = \frac{1}{3}

x = 1.04719 \dots + πn, x = 2.09439 \dots + πn

Grados

x = 6 0^{\circ} + 18 0^{\circ} n, x = 12 0^{\circ} + 18 0^{\circ} n

Pasos de solución

cot^{2} (x) = \frac{1}{3}

Usando el método de sustitución

cot^{2} (x) = \frac{1}{3}

Sea:

cot (x) = u

u^{2} = \frac{1}{3}

u^{2} = \frac{1}{3} : u = \frac{1}{3}, u = - \frac{1}{3}

u^{2} = \frac{1}{3}

Para

x^{2} = f (a)

las soluciones son

x = f (a), - f (a)

u = \frac{1}{3}, u = - \frac{1}{3}

Sustituir en la ecuación

u = cot (x)

cot (x) = \frac{1}{3}, cot (x) = - \frac{1}{3}

cot (x) = \frac{1}{3}, cot (x) = - \frac{1}{3}

cot (x) = \frac{1}{3} : x = arccot (\frac{1}{3}) + πn

cot (x) = \frac{1}{3}

Aplicar propiedades trigonométricas inversas

cot (x) = \frac{1}{3}

Soluciones generales para

cot (x) = \frac{1}{3}

cot (x) = a \Rightarrow x = arccot (a) + πn

x = arccot (\frac{1}{3}) + πn

x = arccot (\frac{1}{3}) + πn

cot (x) = - \frac{1}{3} : x = arccot (- \frac{1}{3}) + πn

cot (x) = - \frac{1}{3}

Aplicar propiedades trigonométricas inversas

cot (x) = - \frac{1}{3}

Soluciones generales para

cot (x) = - \frac{1}{3}

cot (x) = - a \Rightarrow x = arccot (- a) + πn

x = arccot (- \frac{1}{3}) + πn

x = arccot (- \frac{1}{3}) + πn

Combinar toda las soluciones

x = arccot (\frac{1}{3}) + πn, x = arccot (- \frac{1}{3}) + πn

Mostrar soluciones en forma decimal

x = 1.04719 \dots + πn, x = 2.09439 \dots + πn

2 + cos^{2} (x) = - 5 sin (x)

tan^{5} (x) tan^{2} (x) = 1

\frac{sin ^{2} ( a ) + 1}{( 1 + tan ^{2} ( a ) )} = 1

tan (a) = \frac{5}{9}, b = 6

sin^{3} (x) sin^{2} (x) sin (x) = 0