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3tan^2(x)=1,-pi<= x<= pi

Solución

3 tan^{2} (x) = 1, - π \leq x \leq π

Solución

x = \frac{π}{6}, x = - \frac{5 π}{6}, x = - \frac{π}{6}, x = \frac{5 π}{6}

+1

Grados

x = 3 0^{\circ}, x = - 15 0^{\circ}, x = - 3 0^{\circ}, x = 15 0^{\circ}

Pasos de solución

3 tan^{2} (x) = 1, - π \leq x \leq π

Usando el método de sustitución

3 tan^{2} (x) = 1

Sea:

tan (x) = u

3 u^{2} = 1

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

3 u^{2} = 1

Dividir ambos lados entre

3

3 u^{2} = 1

Dividir ambos lados entre

3

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

Simplificar

u^{2} = \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 = tan (x)

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

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

tan (x) = \frac{1}{3}, - π \leq x \leq π : x = \frac{π}{6}, x = - \frac{5 π}{6}

tan (x) = \frac{1}{3}, - π \leq x \leq π

Aplicar propiedades trigonométricas inversas

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

Soluciones generales para

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

tan (x) = a \Rightarrow x = arctan (a) + πn

x = arctan (\frac{1}{3}) + πn

x = arctan (\frac{1}{3}) + πn

Soluciones para el rango

- π \leq x \leq π

x = \frac{π}{6}, x = - \frac{5 π}{6}

tan (x) = - \frac{1}{3}, - π \leq x \leq π : x = - \frac{π}{6}, x = \frac{5 π}{6}

tan (x) = - \frac{1}{3}, - π \leq x \leq π

Aplicar propiedades trigonométricas inversas

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

Soluciones generales para

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

tan (x) = - a \Rightarrow x = arctan (- a) + πn

x = arctan (- \frac{1}{3}) + πn

x = arctan (- \frac{1}{3}) + πn

Soluciones para el rango

- π \leq x \leq π

x = - \frac{π}{6}, x = \frac{5 π}{6}

Combinar toda las soluciones

x = \frac{π}{6}, x = - \frac{5 π}{6}, x = - \frac{π}{6}, x = \frac{5 π}{6}

Ejemplos populares

cos(2t)-cos(t)=0.5

cos (2 t) - cos (t) = 0.5

0=-6csc(x)cot(x)

0 = - 6 csc (x) cot (x)

0.08=0.12cos^2(3.922t)

0.08 = 0.12 cos^{2} (3.922 t)

(15)/(sin(32))=(12)/(sin(b))

\frac{15}{sin ( 3 2 ^{\circ} )} = \frac{12}{sin ( b )}

(sin(x))/(25)=(sin(125))/(165.6)

\frac{sin ( x )}{25} = \frac{sin ( 12 5 ^{\circ} )}{165.6}