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Popular Trigonometria >

-1<tan(x)<1

Solução

- 1 < tan (x) < 1

Solução

πn \leq x < \frac{π}{4} + πn or \frac{3 π}{4} + πn < x < π + πn

+2

Notação de intervalo

[πn, \frac{π}{4} + πn) \cup (\frac{3 π}{4} + πn, π + πn)

Decimal

πn \leq x < 0.78539 \dots + πn or 2.35619 \dots + πn < x < 3.14159 \dots + πn

Passos da solução

- 1 < tan (x) < 1

Se

a < u < b

então

a < u and u < b

- 1 < tan (x) and tan (x) < 1

- 1 < tan (x) : - \frac{π}{4} + πn < x < \frac{π}{2} + πn

- 1 < tan (x)

Trocar lados

tan (x) > - 1

Se

tan (x) > a

então

arctan (a) + πn < x < \frac{π}{2} + πn

arctan (- 1) + πn < x < \frac{π}{2} + πn

Simplificar

arctan (- 1) : - \frac{π}{4}

arctan (- 1)

Utilizar a seguinte propriedade:

arctan (- x) = - arctan (x)

arctan (- 1) = - arctan (1)

= - arctan (1)

Utilizar a seguinte identidade trivial:

arctan (1) = \frac{π}{4}

arctan (1)

x 0 \frac{3}{3} 13 arctan (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} arctan (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ}

= \frac{π}{4}

= - \frac{π}{4}

- \frac{π}{4} + πn < x < \frac{π}{2} + πn

tan (x) < 1 : - \frac{π}{2} + πn < x < \frac{π}{4} + πn

tan (x) < 1

Se

tan (x) < a

então

- \frac{π}{2} + πn < x < arctan (a) + πn

- \frac{π}{2} + πn < x < arctan (1) + πn

Simplificar

arctan (1) : \frac{π}{4}

arctan (1)

Utilizar a seguinte identidade trivial:

arctan (1) = \frac{π}{4}

x 0 \frac{3}{3} 13 arctan (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} arctan (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ}

= \frac{π}{4}

- \frac{π}{2} + πn < x < \frac{π}{4} + πn

Combinar os intervalos

- \frac{π}{4} + πn < x < \frac{π}{2} + πn and - \frac{π}{2} + πn < x < \frac{π}{4} + πn

Junte intervalos que se sobrepoem

πn \leq x < \frac{π}{4} + πn or \frac{3 π}{4} + πn < x < π + πn

Exemplos populares

0<cos(θ)<= sqrt(3)sin(θ)

0 < cos (θ) \leq 3 sin (θ)

sin(t)<0\land cos(t)<0

sin (t) < 0 and cos (t) < 0

csc(x)=-2sqrt(5)\land cos(x)<0,cot(2x)

csc (x) = - 25 and cos (x) < 0, cot (2 x)

sin(θ)= 7/25 \land cos(θ)>0

sin (θ) = \frac{7}{25} and cos (θ) > 0

sin(x)=-4/5 \land cos(x)<0,cos(x/2)

sin (x) = - \frac{4}{5} and cos (x) < 0, cos (\frac{x}{2})