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受欢迎的三角函数 >

(sin(x))^{(sin(x))}= 1/(sqrt(2))

解答

(sin (x))^{(s i n (x))} = \frac{1}{2}

解答

x = 0.25268 \dots + 2 πn, x = π - 0.25268 \dots + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

+1

度数

x = 14.47751 \dots^{\circ} + 36 0^{\circ} n, x = 165.52248 \dots^{\circ} + 36 0^{\circ} n, x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

求解步骤

(sin (x))^{(s i n (x))} = \frac{1}{2}

两边减去

\frac{1}{2}

sin^{s i n (x)} (x) - \frac{1}{2} = 0

化简

sin^{s i n (x)} (x) - \frac{1}{2} : \frac{2 sin ^{s i n (x)} ( x ) - 1}{2}

sin^{s i n (x)} (x) - \frac{1}{2}

将项转换为分式:

sin^{s i n (x)} (x) = \frac{sin ^{s i n (x)} ( x ) 2}{2}

= \frac{sin ^{s i n (x)} ( x ) 2}{2} - \frac{1}{2}

因为分母相等，所以合并分式:

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

= \frac{sin ^{s i n (x)} ( x ) 2 - 1}{2}

\frac{2 sin ^{s i n (x)} ( x ) - 1}{2} = 0

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

2 sin^{s i n (x)} (x) - 1 = 0

使用三角恒等式改写

- 1 + sin^{s i n (x)} (x) 2

使用基本三角恒等式:

sin (x) = \frac{1}{csc ( x )}

= - 1 + (\frac{1}{csc ( x )})^{\frac{1}{c s c ( x )}} 2

- 1 + (\frac{1}{csc ( x )})^{\frac{1}{c s c ( x )}} 2 = 0

用替代法求解

- 1 + (\frac{1}{csc ( x )})^{\frac{1}{c s c ( x )}} 2 = 0

令

: csc (x) = u

- 1 + (\frac{1}{u})^{\frac{1}{u}} 2 = 0

- 1 + (\frac{1}{u})^{\frac{1}{u}} 2 = 0 : u = 4, u = 2

- 1 + (\frac{1}{u})^{\frac{1}{u}} 2 = 0

使用指数运算法则

- 1 + (\frac{1}{u})^{\frac{1}{u}} 2 = 0

使用指数法则:

f (x)^{g (x)} = e^{g (x) l n (f (x))}

(\frac{1}{u})^{\frac{1}{u}} = e^{\frac{1}{u} l n (\frac{1}{u})}

- 1 + e^{\frac{1}{u} l n (\frac{1}{u})} 2 = 0

- 1 + e^{\frac{1}{u} l n (\frac{1}{u})} 2 = 0

两边加上

1

- 1 + e^{\frac{1}{u} l n (\frac{1}{u})} 2 + 1 = 0 + 1

化简

2 e^{\frac{1}{u} l n (\frac{1}{u})} = 1

两边除以

2

2 e^{\frac{1}{u} l n (\frac{1}{u})} = 1

两边除以

2

\frac{2 e ^{\frac{1}{u} l n (\frac{1}{u})}}{2} = \frac{1}{2}

化简

e^{\frac{1}{u} l n (\frac{1}{u})} = \frac{1}{2}

e^{\frac{1}{u} l n (\frac{1}{u})} = \frac{1}{2}

化简

e^{\frac{1}{u} l n (\frac{1}{u})} = \frac{2}{2}

使用指数运算法则

e^{\frac{1}{u} l n (\frac{1}{u})} = \frac{2}{2}

将转换为以

2

为底数的幂:

e^{\frac{1}{u} l n (\frac{1}{u})} = 2^{\frac{- 1}{2}}

化简

e^{\frac{1}{u} l n (\frac{1}{u})} = 2^{\frac{- 1}{2}}

e^{\frac{1}{u} l n (\frac{1}{u})} = 2^{\frac{- 1}{2}}

若

f (x) = g (x)

，则

ln (f (x)) = ln (g (x))

ln (e^{\frac{1}{u} l n (\frac{1}{u})}) = ln (2^{\frac{- 1}{2}})

使用对数计算法则:

ln (e^{a}) = a

ln (e^{\frac{1}{u} l n (\frac{1}{u})}) = \frac{1}{u} ln (\frac{1}{u})

\frac{1}{u} ln (\frac{1}{u}) = ln (2^{\frac{- 1}{2}})

使用对数计算法则:

ln (x^{a}) = a \cdot ln (x)

ln (2^{\frac{- 1}{2}}) = \frac{- 1}{2} ln (2)

\frac{1}{u} ln (\frac{1}{u}) = \frac{- 1}{2} ln (2)

\frac{1}{u} ln (\frac{1}{u}) = \frac{- 1}{2} ln (2)

解

\frac{1}{u} ln (\frac{1}{u}) = \frac{- 1}{2} ln (2) : u = 4, u = 2

\frac{1}{u} ln (\frac{1}{u}) = \frac{- 1}{2} ln (2)

在两边乘以

u

\frac{1}{u} ln (\frac{1}{u}) = \frac{- 1}{2} ln (2)

在两边乘以

u

\frac{1}{u} ln (\frac{1}{u}) u = \frac{- 1}{2} ln (2) u

化简

\frac{1}{u} ln (\frac{1}{u}) u = \frac{- 1}{2} ln (2) u

化简

\frac{1}{u} ln (\frac{1}{u}) u : ln (\frac{1}{u})

\frac{1}{u} ln (\frac{1}{u}) u

分式相乘:

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

= \frac{1 \cdot ln ( \frac{1}{u} ) u}{u}

约分:

u

= 1 \cdot ln (\frac{1}{u})

乘以:

1 \cdot ln (\frac{1}{u}) = ln (\frac{1}{u})

= ln (\frac{1}{u})

化简

\frac{- 1}{2} ln (2) u : - \frac{1}{2} u ln (2)

\frac{- 1}{2} ln (2) u

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{1}{2} u ln (2)

ln (\frac{1}{u}) = - \frac{1}{2} u ln (2)

ln (\frac{1}{u}) = - \frac{1}{2} u ln (2)

ln (\frac{1}{u}) = - \frac{1}{2} u ln (2)

若

f (x) = g (x)

，则

a^{f (x)} = a^{g (x)}

e^{l n (\frac{1}{u})} = e^{- \frac{1}{2} u l n (2)}

化简

e^{l n (\frac{1}{u})} : \frac{1}{u}

e^{l n (\frac{1}{u})}

使用对数计算法则:

a^{l o g_{a} (b)} = b

= \frac{1}{u}

化简

e^{- \frac{1}{2} u l n (2)} : 2^{- \frac{1}{2} u}

e^{- \frac{1}{2} u l n (2)}

使用指数法则:

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

= (e^{l n (2)})^{- \frac{1}{2} u}

使用对数计算法则:

a^{l o g_{a} (b)} = b

e^{l n (2)} = 2

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

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

在两边乘以

u

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

化简

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

解

1 = 2^{- \frac{1}{2} u} u : u = 4, u = 2

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

将

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

调整为朗伯形式:

1 = e^{- \frac{1}{2} l n (2) u} u

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

x e^{x} = a

是朗伯形式的方程

使用指数运算法则

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

将转换为以

e

为底数的幂:

1 = e^{l n (2) (- \frac{1}{2} u)} u

使用指数法则:

a = b^{l o g_{b} (a)}

2^{- \frac{1}{2} u} = (e^{l n (2)})^{- \frac{1}{2} u}

1 = (e^{l n (2)})^{- \frac{1}{2} u} u

使用指数法则:

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

(e^{l n (2)})^{- \frac{1}{2} u} = e^{l n (2) (- \frac{1}{2} u)}

1 = e^{l n (2) (- \frac{1}{2} u)} u

1 = e^{l n (2) (- \frac{1}{2} u)} u

化简

1 = e^{- \frac{1}{2} l n (2) u} u

1 = e^{- \frac{1}{2} l n (2) u} u

用

- \frac{1}{2} u ln (2) = v

和

u = - \frac{2 v}{ln ( 2 )}

改写方程式

1 = e^{v} (- \frac{2 v}{ln ( 2 )})

改写

1 = e^{v} (- \frac{2 v}{ln ( 2 )})

为朗伯形式:

e^{v} v = - \frac{ln ( 2 )}{2}

1 = e^{v} (- \frac{2 v}{ln ( 2 )})

x e^{x} = a

是朗伯形式的方程

交换两边

e^{v} (- \frac{2 v}{ln ( 2 )}) = 1

在两边乘以

ln (2)

e^{v} (- \frac{2 v}{ln ( 2 )}) ln (2) = 1 \cdot ln (2)

化简

- 2 e^{v} v = ln (2)

两边除以

- 2

\frac{- 2 e ^{v} v}{- 2} = \frac{ln ( 2 )}{- 2}

化简

e^{v} v = - \frac{ln ( 2 )}{2}

解

e^{v} v = - \frac{ln ( 2 )}{2} : v = - 2 ln (2), v = - ln (2)

e^{v} v = - \frac{ln ( 2 )}{2}

x e^{x} = a

（

- \frac{1}{e} \leq a < 0

）的解为朗伯

W

函数的主分支和负分支:

x = W_{0} (a), W_{- 1} (a)

v = W_{- 1} (- \frac{ln ( 2 )}{2}), v = W_{0} (- \frac{ln ( 2 )}{2})

化简

v = - 2 ln (2), v = - ln (2)

v = - 2 ln (2), v = - ln (2)

代回

v = - \frac{1}{2} u ln (2)

，求解

u

解

- \frac{1}{2} u ln (2) = - 2 ln (2) : u = 4

- \frac{1}{2} u ln (2) = - 2 ln (2)

在两边乘以

- 2

- \frac{1}{2} u ln (2) = - 2 ln (2)

在两边乘以

- 2

(- \frac{1}{2} u ln (2)) (- 2) = (- 2 ln (2)) (- 2)

化简

ln (2) u = 4 ln (2)

ln (2) u = 4 ln (2)

两边除以

ln (2)

ln (2) u = 4 ln (2)

两边除以

ln (2)

\frac{ln ( 2 ) u}{ln ( 2 )} = \frac{4 ln ( 2 )}{ln ( 2 )}

化简

u = 4

u = 4

解

- \frac{1}{2} u ln (2) = - ln (2) : u = 2

- \frac{1}{2} u ln (2) = - ln (2)

在两边乘以

- 2

- \frac{1}{2} u ln (2) = - ln (2)

在两边乘以

- 2

(- \frac{1}{2} u ln (2)) (- 2) = (- ln (2)) (- 2)

化简

ln (2) u = 2 ln (2)

ln (2) u = 2 ln (2)

两边除以

ln (2)

ln (2) u = 2 ln (2)

两边除以

ln (2)

\frac{ln ( 2 ) u}{ln ( 2 )} = \frac{2 ln ( 2 )}{ln ( 2 )}

化简

u = 2

u = 2

u = 4, u = 2

u = 4, u = 2

验证解:

u = 4

真

, u = 2

真

将它们代入

\frac{1}{u} ln (\frac{1}{u}) = \frac{- 1}{2} ln (2)

检验解是否符合

去除与方程不符的解。

代入

u = 4 :

真

\frac{1}{4} ln (\frac{1}{4}) = \frac{- 1}{2} ln (2)

\frac{1}{4} ln (\frac{1}{4}) = - \frac{1}{2} ln (2)

\frac{1}{4} ln (\frac{1}{4})

化简

ln (\frac{1}{4}) : - 2 ln (2)

ln (\frac{1}{4})

使用对数计算法则:

lo g_{a} (\frac{1}{x}) = - lo g_{a} (x)

= - ln (4)

改写

4

为幂形式:

4 = 2^{2}

= - ln (2^{2})

使用对数计算法则:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

ln (2^{2}) = 2 ln (2)

= - 2 ln (2)

= \frac{1}{4} (- 2 ln (2))

去除括号:

(- a) = - a

= - \frac{1}{4} \cdot 2 ln (2)

分式相乘:

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

= - \frac{1 \cdot 2}{4} ln (2)

\frac{1 \cdot 2}{4} = \frac{1}{2}

\frac{1 \cdot 2}{4}

数字相乘:

1 \cdot 2 = 2

= \frac{2}{4}

约分:

2

= \frac{1}{2}

= - \frac{1}{2} ln (2)

\frac{- 1}{2} ln (2) = - \frac{1}{2} ln (2)

\frac{- 1}{2} ln (2)

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{1}{2} ln (2)

- \frac{1}{2} ln (2) = - \frac{1}{2} ln (2)

真

代入

u = 2 :

真

\frac{1}{2} ln (\frac{1}{2}) = \frac{- 1}{2} ln (2)

\frac{1}{2} ln (\frac{1}{2}) = - \frac{1}{2} ln (2)

\frac{1}{2} ln (\frac{1}{2})

化简

ln (\frac{1}{2}) : - ln (2)

ln (\frac{1}{2})

使用对数计算法则:

lo g_{a} (\frac{1}{x}) = - lo g_{a} (x)

= - ln (2)

= \frac{1}{2} (- ln (2))

去除括号:

(- a) = - a

= - \frac{1}{2} ln (2)

\frac{- 1}{2} ln (2) = - \frac{1}{2} ln (2)

\frac{- 1}{2} ln (2)

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{1}{2} ln (2)

- \frac{1}{2} ln (2) = - \frac{1}{2} ln (2)

真

解为

u = 4, u = 2

u = 4, u = 2

u = csc (x)

代回

csc (x) = 4, csc (x) = 2

csc (x) = 4, csc (x) = 2

csc (x) = 4 : x = arccsc (4) + 2 πn, x = π - arccsc (4) + 2 πn

csc (x) = 4

使用反三角函数性质

csc (x) = 4

csc (x) = 4

的通解

csc (x) = a \Rightarrow x = arccsc (a) + 2 πn, x = π - arccsc (a) + 2 πn

x = arccsc (4) + 2 πn, x = π - arccsc (4) + 2 πn

x = arccsc (4) + 2 πn, x = π - arccsc (4) + 2 πn

csc (x) = 2 : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

csc (x) = 2

csc (x) = 2

的通解

csc (x)

周期表（周期为

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} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 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} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

合并所有解

x = arccsc (4) + 2 πn, x = π - arccsc (4) + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

以小数形式表示解

x = 0.25268 \dots + 2 πn, x = π - 0.25268 \dots + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

作图

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