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受欢迎的三角函数 >

sin((x*pi}{(\frac{1+sqrt(5))/2)^2})>0

解答

sin \frac{x \cdot π}{( \frac{1 + 5}{2} ) ^{2}} > 0

解答

(3 + 5) n < x < \frac{3 + 5}{2} + (3 + 5) n

+2

间隔符号

((3 + 5) n, \frac{3 + 5}{2} + (3 + 5) n)

十进制

(3 + 5) n < x < 2.61803 \dots + (3 + 5) n

求解步骤

sin \frac{x π}{( \frac{1 + 5}{2} ) ^{2}} > 0

对于

sin (x) > a

，若

- 1 \leq a < 1

，则

arcsin (a) + 2 πn < x < π - arcsin (a) + 2 πn

arcsin (0) + 2 πn < \frac{x π}{( \frac{1 + 5}{2} ) ^{2}} < π - arcsin (0) + 2 πn

若

a < u < b

，则

a < u and u < b

arcsin (0) + 2 πn < \frac{x π}{( \frac{1 + 5}{2} ) ^{2}} and \frac{x π}{( \frac{1 + 5}{2} ) ^{2}} < π - arcsin (0) + 2 πn

arcsin (0) + 2 πn < \frac{x π}{( \frac{1 + 5}{2} ) ^{2}} : x > (3 + 5) n

arcsin (0) + 2 πn < \frac{x π}{( \frac{1 + 5}{2} ) ^{2}}

交换两边

\frac{x π}{( \frac{1 + 5}{2} ) ^{2}} > arcsin (0) + 2 πn

化简

arcsin (0) + 2 πn : 2 πn

arcsin (0) + 2 πn

使用以下普通恒等式:

arcsin (0) = 0

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= 0 + 2 πn

0 + 2 πn = 2 πn

= 2 πn

\frac{x π}{( \frac{1 + 5}{2} ) ^{2}} > 2 πn

在两边乘以

(\frac{1 + 5}{2})^{2}

\frac{x π}{( \frac{1 + 5}{2} ) ^{2}} > 2 πn

在两边乘以

(\frac{1 + 5}{2})^{2}

\frac{x π ( \frac{1 + 5}{2} ) ^{2}}{( \frac{1 + 5}{2} ) ^{2}} > 2 πn (\frac{1 + 5}{2})^{2}

化简

\frac{x π ( \frac{1 + 5}{2} ) ^{2}}{( \frac{1 + 5}{2} ) ^{2}} > 2 πn (\frac{1 + 5}{2})^{2}

化简

\frac{x π ( \frac{1 + 5}{2} ) ^{2}}{( \frac{1 + 5}{2} ) ^{2}} : π x

\frac{x π ( \frac{1 + 5}{2} ) ^{2}}{( \frac{1 + 5}{2} ) ^{2}}

约分:

(\frac{1 + 5}{2})^{2}

= π x

化简

2 πn (\frac{1 + 5}{2})^{2} : π (3 + 5) n

2 πn (\frac{1 + 5}{2})^{2}

(\frac{1 + 5}{2})^{2} = \frac{3 + 5}{2}

(\frac{1 + 5}{2})^{2}

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{( 1 + 5 ) ^{2}}{2 ^{2}}

(1 + 5)^{2} = 6 + 25

(1 + 5)^{2}

使用完全平方公式:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = 1, b = 5

= 1^{2} + 2 \cdot 1 \cdot 5 + (5)^{2}

化简

1^{2} + 2 \cdot 1 \cdot 5 + (5)^{2} : 6 + 25

1^{2} + 2 \cdot 1 \cdot 5 + (5)^{2}

使用法则

1^{a} = 1

1^{2} = 1

= 1 + 2 \cdot 1 \cdot 5 + (5)^{2}

2 \cdot 1 \cdot 5 = 25

2 \cdot 1 \cdot 5

数字相乘:

2 \cdot 1 = 2

= 25

(5)^{2} = 5

(5)^{2}

使用根式运算法则:

a = a^{\frac{1}{2}}

= (5^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= 5^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

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

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 5

= 1 + 25 + 5

数字相加:

1 + 5 = 6

= 6 + 25

= 6 + 25

= \frac{6 + 2 5}{2 ^{2}}

分解

6 + 25 : 2 (3 + 5)

6 + 25

改写为

= 2 \cdot 3 + 25

因式分解出通项

2

= 2 (3 + 5)

= \frac{2 ( 3 + 5 )}{2 ^{2}}

约分:

2

= \frac{3 + 5}{2}

= 2 π \frac{3 + 5}{2} n

分式相乘:

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

= \frac{2 π ( 3 + 5 )}{2} n

约分:

2

= n (3 + 5) π

π x > π (3 + 5) n

π x > π (3 + 5) n

π x > π (3 + 5) n

两边除以

π

π x > π (3 + 5) n

两边除以

π

\frac{π x}{π} > \frac{π ( 3 + 5 ) n}{π}

化简

x > (3 + 5) n

x > (3 + 5) n

\frac{x π}{( \frac{1 + 5}{2} ) ^{2}} < π - arcsin (0) + 2 πn : x < \frac{3 + 5}{2} + (3 + 5) n

\frac{x π}{( \frac{1 + 5}{2} ) ^{2}} < π - arcsin (0) + 2 πn

化简

π - arcsin (0) + 2 πn : π + 2 πn

π - arcsin (0) + 2 πn

使用以下普通恒等式:

arcsin (0) = 0

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= π - 0 + 2 πn

π - 0 + 2 πn = π + 2 πn

= π + 2 πn

\frac{x π}{( \frac{1 + 5}{2} ) ^{2}} < π + 2 πn

在两边乘以

(\frac{1 + 5}{2})^{2}

\frac{x π}{( \frac{1 + 5}{2} ) ^{2}} < π + 2 πn

在两边乘以

(\frac{1 + 5}{2})^{2}

\frac{x π ( \frac{1 + 5}{2} ) ^{2}}{( \frac{1 + 5}{2} ) ^{2}} < π (\frac{1 + 5}{2})^{2} + 2 πn (\frac{1 + 5}{2})^{2}

化简

\frac{x π ( \frac{1 + 5}{2} ) ^{2}}{( \frac{1 + 5}{2} ) ^{2}} < π (\frac{1 + 5}{2})^{2} + 2 πn (\frac{1 + 5}{2})^{2}

化简

\frac{x π ( \frac{1 + 5}{2} ) ^{2}}{( \frac{1 + 5}{2} ) ^{2}} : π x

\frac{x π ( \frac{1 + 5}{2} ) ^{2}}{( \frac{1 + 5}{2} ) ^{2}}

约分:

(\frac{1 + 5}{2})^{2}

= π x

化简

π (\frac{1 + 5}{2})^{2} + 2 πn (\frac{1 + 5}{2})^{2} : \frac{π ( 3 + 5 )}{2} + π (3 + 5) n

π (\frac{1 + 5}{2})^{2} + 2 πn (\frac{1 + 5}{2})^{2}

π (\frac{1 + 5}{2})^{2} = \frac{π ( 3 + 5 )}{2}

π (\frac{1 + 5}{2})^{2}

(\frac{1 + 5}{2})^{2} = \frac{3 + 5}{2}

(\frac{1 + 5}{2})^{2}

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{( 1 + 5 ) ^{2}}{2 ^{2}}

(1 + 5)^{2} = 6 + 25

(1 + 5)^{2}

使用完全平方公式:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = 1, b = 5

= 1^{2} + 2 \cdot 1 \cdot 5 + (5)^{2}

化简

1^{2} + 2 \cdot 1 \cdot 5 + (5)^{2} : 6 + 25

1^{2} + 2 \cdot 1 \cdot 5 + (5)^{2}

使用法则

1^{a} = 1

1^{2} = 1

= 1 + 2 \cdot 1 \cdot 5 + (5)^{2}

2 \cdot 1 \cdot 5 = 25

2 \cdot 1 \cdot 5

数字相乘:

2 \cdot 1 = 2

= 25

(5)^{2} = 5

(5)^{2}

使用根式运算法则:

a = a^{\frac{1}{2}}

= (5^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= 5^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

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

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 5

= 1 + 25 + 5

数字相加:

1 + 5 = 6

= 6 + 25

= 6 + 25

= \frac{6 + 2 5}{2 ^{2}}

分解

6 + 25 : 2 (3 + 5)

6 + 25

改写为

= 2 \cdot 3 + 25

因式分解出通项

2

= 2 (3 + 5)

= \frac{2 ( 3 + 5 )}{2 ^{2}}

约分:

2

= \frac{3 + 5}{2}

= π \frac{3 + 5}{2}

分式相乘:

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

= \frac{( 3 + 5 ) π}{2}

2 πn (\frac{1 + 5}{2})^{2} = π (3 + 5) n

2 πn (\frac{1 + 5}{2})^{2}

(\frac{1 + 5}{2})^{2} = \frac{3 + 5}{2}

(\frac{1 + 5}{2})^{2}

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{( 1 + 5 ) ^{2}}{2 ^{2}}

(1 + 5)^{2} = 6 + 25

(1 + 5)^{2}

使用完全平方公式:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = 1, b = 5

= 1^{2} + 2 \cdot 1 \cdot 5 + (5)^{2}

化简

1^{2} + 2 \cdot 1 \cdot 5 + (5)^{2} : 6 + 25

1^{2} + 2 \cdot 1 \cdot 5 + (5)^{2}

使用法则

1^{a} = 1

1^{2} = 1

= 1 + 2 \cdot 1 \cdot 5 + (5)^{2}

2 \cdot 1 \cdot 5 = 25

2 \cdot 1 \cdot 5

数字相乘:

2 \cdot 1 = 2

= 25

(5)^{2} = 5

(5)^{2}

使用根式运算法则:

a = a^{\frac{1}{2}}

= (5^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= 5^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

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

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 5

= 1 + 25 + 5

数字相加:

1 + 5 = 6

= 6 + 25

= 6 + 25

= \frac{6 + 2 5}{2 ^{2}}

分解

6 + 25 : 2 (3 + 5)

6 + 25

改写为

= 2 \cdot 3 + 25

因式分解出通项

2

= 2 (3 + 5)

= \frac{2 ( 3 + 5 )}{2 ^{2}}

约分:

2

= \frac{3 + 5}{2}

= 2 π \frac{3 + 5}{2} n

分式相乘:

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

= \frac{2 π ( 3 + 5 )}{2} n

约分:

2

= n (3 + 5) π

= \frac{π ( 3 + 5 )}{2} + π (3 + 5) n

π x < \frac{π ( 3 + 5 )}{2} + π (3 + 5) n

π x < \frac{π ( 3 + 5 )}{2} + π (3 + 5) n

π x < \frac{π ( 3 + 5 )}{2} + π (3 + 5) n

两边除以

π

π x < \frac{π ( 3 + 5 )}{2} + π (3 + 5) n

两边除以

π

\frac{π x}{π} < \frac{\frac{π ( 3 + 5 )}{2}}{π} + \frac{π ( 3 + 5 ) n}{π}

化简

\frac{π x}{π} < \frac{\frac{π ( 3 + 5 )}{2}}{π} + \frac{π ( 3 + 5 ) n}{π}

化简

\frac{π x}{π} : x

\frac{π x}{π}

约分:

π

= x

化简

\frac{\frac{π ( 3 + 5 )}{2}}{π} + \frac{π ( 3 + 5 ) n}{π} : \frac{3 + 5}{2} + (3 + 5) n

\frac{\frac{π ( 3 + 5 )}{2}}{π} + \frac{π ( 3 + 5 ) n}{π}

\frac{\frac{π ( 3 + 5 )}{2}}{π} = \frac{3 + 5}{2}

\frac{\frac{π ( 3 + 5 )}{2}}{π}

使用分式法则:

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

= \frac{π ( 3 + 5 )}{2 π}

约分:

π

= \frac{3 + 5}{2}

\frac{π ( 3 + 5 ) n}{π} = (3 + 5) n

\frac{π ( 3 + 5 ) n}{π}

约分:

π

= (3 + 5) n

= \frac{3 + 5}{2} + (3 + 5) n

x < \frac{3 + 5}{2} + (3 + 5) n

x < \frac{3 + 5}{2} + (3 + 5) n

x < \frac{3 + 5}{2} + (3 + 5) n

合并区间

x > (3 + 5) n and x < \frac{3 + 5}{2} + (3 + 5) n

合并重叠的区间

(3 + 5) n < x < \frac{3 + 5}{2} + (3 + 5) n

流行的例子

2sin(x)-sqrt(3)<= 0

2 sin (x) - 3 \leq 0

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

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

(2+sqrt(3)sin(x))<0

(2 + 3 sin (x)) < 0

2 cos (x) - 1 > 0

sin(2x)-cos(2x)>0

sin (2 x) - cos (2 x) > 0