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Popular Trigonometria >

(2sin(2x)+sqrt(2))*tan(x)<0

Solução

(2 sin (2 x) + 2) \cdot tan (x) < 0

Solução

\frac{π}{2} + πn < x < \frac{5 π}{8} + πn or \frac{7 π}{8} + πn < x < π + πn

Notação de intervalo

(\frac{π}{2} + πn, \frac{5 π}{8} + πn) \cup (\frac{7 π}{8} + πn, π + πn)

Decimal

1.57079 \dots + πn < x < 1.96349 \dots + πn or 2.74889 \dots + πn < x < 3.14159 \dots + πn

Passos da solução

(2 sin (2 x) + 2) tan (x) < 0

Periodicidade de

(2 sin (2 x) + 2) tan (x) : π

(2 sin (2 x) + 2) tan (x)

é composta pelas seguintes funções e períodos:

sin (2 x)

com periodicidade de

\frac{2 π}{2}

A periodicidade composta é:

= π

Expresar com seno, cosseno

(2 sin (2 x) + 2) tan (x) < 0

Utilizar a Identidade Básica da Trigonometria:

tan (x) = \frac{sin ( x )}{cos ( x )}

(2 sin (2 x) + 2) \frac{sin ( x )}{cos ( x )} < 0

(2 sin (2 x) + 2) \frac{sin ( x )}{cos ( x )} < 0

Simplificar

(2 sin (2 x) + 2) \frac{sin ( x )}{cos ( x )} : \frac{sin ( x ) ( 2 sin ( 2 x ) + 2 )}{cos ( x )}

(2 sin (2 x) + 2) \frac{sin ( x )}{cos ( x )}

Multiplicar frações:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{sin ( x ) ( 2 sin ( 2 x ) + 2 )}{cos ( x )}

\frac{sin ( x ) ( 2 sin ( 2 x ) + 2 )}{cos ( x )} < 0

Encontre os zeros e pontos indefinidos de

\frac{sin ( x ) ( 2 sin ( 2 x ) + 2 )}{cos ( x )}

para

0 \leq x < π

Para encontrar os zeros, defina a desigualdade como zero

\frac{sin ( x ) ( 2 sin ( 2 x ) + 2 )}{cos ( x )} = 0

\frac{sin ( x ) ( 2 sin ( 2 x ) + 2 )}{cos ( x )} = 0, 0 \leq x < π : x = 0, x = \frac{5 π}{8}, x = \frac{7 π}{8}

\frac{sin ( x ) ( 2 sin ( 2 x ) + 2 )}{cos ( x )} = 0, 0 \leq x < π

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

sin (x) (2 sin (2 x) + 2) = 0

Resolver cada parte separadamente

sin (x) = 0 or 2 sin (2 x) + 2 = 0

sin (x) = 0, 0 \leq x < π : x = 0

sin (x) = 0, 0 \leq x < π

Soluções gerais para

sin (x) = 0

sin (x)

tabela de periodicidade com ciclo de

2 πn

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Resolver

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

Soluções para o intervalo

0 \leq x < π

x = 0

2 sin (2 x) + 2 = 0, 0 \leq x < π : x = \frac{5 π}{8}, x = \frac{7 π}{8}

2 sin (2 x) + 2 = 0, 0 \leq x < π

Mova

2

para o lado direito

2 sin (2 x) + 2 = 0

Subtrair

2

de ambos os lados

2 sin (2 x) + 2 - 2 = 0 - 2

Simplificar

2 sin (2 x) = - 2

2 sin (2 x) = - 2

Dividir ambos os lados por

2

2 sin (2 x) = - 2

Dividir ambos os lados por

2

\frac{2 sin ( 2 x )}{2} = \frac{- 2}{2}

Simplificar

sin (2 x) = - \frac{2}{2}

sin (2 x) = - \frac{2}{2}

Soluções gerais para

sin (2 x) = - \frac{2}{2}

sin (x)

tabela de periodicidade com ciclo de

2 πn

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

2 x = \frac{5 π}{4} + 2 πn, 2 x = \frac{7 π}{4} + 2 πn

2 x = \frac{5 π}{4} + 2 πn, 2 x = \frac{7 π}{4} + 2 πn

Resolver

2 x = \frac{5 π}{4} + 2 πn : x = \frac{5 π}{8} + πn

2 x = \frac{5 π}{4} + 2 πn

Dividir ambos os lados por

2

2 x = \frac{5 π}{4} + 2 πn

Dividir ambos os lados por

2

\frac{2 x}{2} = \frac{\frac{5 π}{4}}{2} + \frac{2 πn}{2}

Simplificar

\frac{2 x}{2} = \frac{\frac{5 π}{4}}{2} + \frac{2 πn}{2}

Simplificar

\frac{2 x}{2} : x

\frac{2 x}{2}

Dividir:

\frac{2}{2} = 1

= x

Simplificar

\frac{\frac{5 π}{4}}{2} + \frac{2 πn}{2} : \frac{5 π}{8} + πn

\frac{\frac{5 π}{4}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{4}}{2} = \frac{5 π}{8}

\frac{\frac{5 π}{4}}{2}

Aplicar as propriedades das frações:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{5 π}{4 \cdot 2}

Multiplicar os números:

4 \cdot 2 = 8

= \frac{5 π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Dividir:

\frac{2}{2} = 1

= πn

= \frac{5 π}{8} + πn

x = \frac{5 π}{8} + πn

x = \frac{5 π}{8} + πn

x = \frac{5 π}{8} + πn

Resolver

2 x = \frac{7 π}{4} + 2 πn : x = \frac{7 π}{8} + πn

2 x = \frac{7 π}{4} + 2 πn

Dividir ambos os lados por

2

2 x = \frac{7 π}{4} + 2 πn

Dividir ambos os lados por

2

\frac{2 x}{2} = \frac{\frac{7 π}{4}}{2} + \frac{2 πn}{2}

Simplificar

\frac{2 x}{2} = \frac{\frac{7 π}{4}}{2} + \frac{2 πn}{2}

Simplificar

\frac{2 x}{2} : x

\frac{2 x}{2}

Dividir:

\frac{2}{2} = 1

= x

Simplificar

\frac{\frac{7 π}{4}}{2} + \frac{2 πn}{2} : \frac{7 π}{8} + πn

\frac{\frac{7 π}{4}}{2} + \frac{2 πn}{2}

\frac{\frac{7 π}{4}}{2} = \frac{7 π}{8}

\frac{\frac{7 π}{4}}{2}

Aplicar as propriedades das frações:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{7 π}{4 \cdot 2}

Multiplicar os números:

4 \cdot 2 = 8

= \frac{7 π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Dividir:

\frac{2}{2} = 1

= πn

= \frac{7 π}{8} + πn

x = \frac{7 π}{8} + πn

x = \frac{7 π}{8} + πn

x = \frac{7 π}{8} + πn

x = \frac{5 π}{8} + πn, x = \frac{7 π}{8} + πn

Soluções para o intervalo

0 \leq x < π

x = \frac{5 π}{8}, x = \frac{7 π}{8}

Combinar toda as soluções

x = 0, x = \frac{5 π}{8}, x = \frac{7 π}{8}

Encontre os pontos indefinidos:

x = \frac{π}{2}

Encontre os zeros do denominador

cos (x) = 0

Soluções gerais para

cos (x) = 0

cos (x)

tabela de periodicidade com ciclo de

2 πn

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Soluções para o intervalo

0 \leq x < π

x = \frac{π}{2}

0, \frac{π}{2}, \frac{5 π}{8}, \frac{7 π}{8}

Identifique os intervalos

0 < x < \frac{π}{2}, \frac{π}{2} < x < \frac{5 π}{8}, \frac{5 π}{8} < x < \frac{7 π}{8}, \frac{7 π}{8} < x < π

Resumir em uma tabela:

sin (x) 2 sin (2 x) + 2 cos (x) \frac{s i n ( x ) ( 2 s i n ( 2 x ) + 2 )}{c o s ( x )} x = 0 0 + + 0 0 < x < \frac{π}{2} + + + + x = \frac{π}{2} + + 0 Indefinido \frac{π}{2} < x < \frac{5 π}{8} + + - - x = \frac{5 π}{8} + 0 - 0 \frac{5 π}{8} < x < \frac{7 π}{8} + - - + x = \frac{7 π}{8} + 0 - 0 \frac{7 π}{8} < x < π + + - - x = π 0 + - 0

Identifique os intervalos que satisfaçam à condição necessária:

< 0

\frac{π}{2} < x < \frac{5 π}{8} or \frac{7 π}{8} < x < π

Utilizar a periodicidade de

(2 sin (2 x) + 2) tan (x)

\frac{π}{2} + πn < x < \frac{5 π}{8} + πn or \frac{7 π}{8} + πn < x < π + πn

Exemplos populares

cos(x)<=-(sqrt(2))/2 ,-pi<= x<= pi

cos (x) \leq - \frac{2}{2}, - π \leq x \leq π

6sin(2x-(2pi)/3)>0

6 sin (2 x - \frac{2 π}{3}) > 0

1>tan(x)

1 > tan (x)

cos(x)-(sqrt(3))/2 <= 0

cos (x) - \frac{3}{2} \leq 0

tan(x)<-\sqrt[4]{5},-pi<= x<= pi

tan (x) < - 45, - π \leq x \leq π