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受欢迎的三角函数 >

tan(2θ)+6tan(θ)+8=0

解答

tan (2 θ) + 6 tan (θ) + 8 = 0

解答

θ = - 0.64552 \dots + πn, θ = 0.81999 \dots + πn, θ = - 1.02643 \dots + πn

+1

度数

θ = - 36.98596 \dots^{\circ} + 18 0^{\circ} n, θ = 46.98210 \dots^{\circ} + 18 0^{\circ} n, θ = - 58.81021 \dots^{\circ} + 18 0^{\circ} n

求解步骤

tan (2 θ) + 6 tan (θ) + 8 = 0

使用三角恒等式改写

8 + tan (2 θ) + 6 tan (θ)

使用倍角公式:

tan (2 x) = \frac{2 tan ( x )}{1 - tan ^{2} ( x )}

= 8 + \frac{2 tan ( θ )}{1 - tan ^{2} ( θ )} + 6 tan (θ)

合并分式

\frac{2 tan ( θ )}{- tan ^{2} ( θ ) + 1} + 6 tan (θ) : \frac{8 tan ( θ ) - 6 tan ^{3} ( θ )}{1 - tan ^{2} ( θ )}

\frac{2 tan ( θ )}{- tan ^{2} ( θ ) + 1} + 6 tan (θ)

将项转换为分式:

6 tan (θ) = \frac{6 tan ( θ ) ( 1 - tan ^{2} ( θ ) )}{1 - tan ^{2} ( θ )}

= \frac{2 tan ( θ )}{1 - tan ^{2} ( θ )} + \frac{6 tan ( θ ) ( 1 - tan ^{2} ( θ ) )}{1 - tan ^{2} ( θ )}

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{2 tan ( θ ) + 6 tan ( θ ) ( 1 - tan ^{2} ( θ ) )}{1 - tan ^{2} ( θ )}

乘开

2 tan (θ) + 6 tan (θ) (1 - tan^{2} (θ)) : 8 tan (θ) - 6 tan^{3} (θ)

2 tan (θ) + 6 tan (θ) (1 - tan^{2} (θ))

乘开

6 tan (θ) (1 - tan^{2} (θ)) : 6 tan (θ) - 6 tan^{3} (θ)

6 tan (θ) (1 - tan^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = 6 tan (θ), b = 1, c = tan^{2} (θ)

= 6 tan (θ) \cdot 1 - 6 tan (θ) tan^{2} (θ)

= 6 \cdot 1 \cdot tan (θ) - 6 tan^{2} (θ) tan (θ)

化简

6 \cdot 1 \cdot tan (θ) - 6 tan^{2} (θ) tan (θ) : 6 tan (θ) - 6 tan^{3} (θ)

6 \cdot 1 \cdot tan (θ) - 6 tan^{2} (θ) tan (θ)

6 \cdot 1 \cdot tan (θ) = 6 tan (θ)

6 \cdot 1 \cdot tan (θ)

数字相乘:

6 \cdot 1 = 6

= 6 tan (θ)

6 tan^{2} (θ) tan (θ) = 6 tan^{3} (θ)

6 tan^{2} (θ) tan (θ)

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

tan^{2} (θ) tan (θ) = tan^{2 + 1} (θ)

= 6 tan^{2 + 1} (θ)

数字相加:

2 + 1 = 3

= 6 tan^{3} (θ)

= 6 tan (θ) - 6 tan^{3} (θ)

= 6 tan (θ) - 6 tan^{3} (θ)

= 2 tan (θ) + 6 tan (θ) - 6 tan^{3} (θ)

同类项相加:

2 tan (θ) + 6 tan (θ) = 8 tan (θ)

= 8 tan (θ) - 6 tan^{3} (θ)

= \frac{8 tan ( θ ) - 6 tan ^{3} ( θ )}{1 - tan ^{2} ( θ )}

= \frac{8 tan ( θ ) - 6 tan ^{3} ( θ )}{1 - tan ^{2} ( θ )} + 8

8 + \frac{- 6 tan ^{3} ( θ ) + 8 tan ( θ )}{1 - tan ^{2} ( θ )} = 0

8 + \frac{- 6 tan ^{3} ( θ ) + 8 tan ( θ )}{1 - tan ^{2} ( θ )} = 0

用替代法求解

8 + \frac{- 6 tan ^{3} ( θ ) + 8 tan ( θ )}{1 - tan ^{2} ( θ )} = 0

令

: tan (θ) = u

8 + \frac{- 6 u ^{3} + 8 u}{1 - u ^{2}} = 0

8 + \frac{- 6 u ^{3} + 8 u}{1 - u ^{2}} = 0 : u \approx - 0.75317 \dots, u \approx 1.07169 \dots, u \approx - 1.65186 \dots

8 + \frac{- 6 u ^{3} + 8 u}{1 - u ^{2}} = 0

在两边乘以

1 - u^{2}

8 + \frac{- 6 u ^{3} + 8 u}{1 - u ^{2}} = 0

在两边乘以

1 - u^{2}

8 (1 - u^{2}) + \frac{- 6 u ^{3} + 8 u}{1 - u ^{2}} (1 - u^{2}) = 0 \cdot (1 - u^{2})

化简

8 (1 - u^{2}) + \frac{- 6 u ^{3} + 8 u}{1 - u ^{2}} (1 - u^{2}) = 0 \cdot (1 - u^{2})

化简

\frac{- 6 u ^{3} + 8 u}{1 - u ^{2}} (1 - u^{2}) : - 6 u^{3} + 8 u

\frac{- 6 u ^{3} + 8 u}{1 - u ^{2}} (1 - u^{2})

分式相乘:

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

= \frac{( - 6 u ^{3} + 8 u ) ( 1 - u ^{2} )}{1 - u ^{2}}

约分:

1 - u^{2}

= - - 6 u^{3} + 8 u

化简

0 \cdot (1 - u^{2}) : 0

0 \cdot (1 - u^{2})

使用法则

0 \cdot a = 0

= 0

8 (1 - u^{2}) - 6 u^{3} + 8 u = 0

8 (1 - u^{2}) - 6 u^{3} + 8 u = 0

8 (1 - u^{2}) - 6 u^{3} + 8 u = 0

解

8 (1 - u^{2}) - 6 u^{3} + 8 u = 0 : u \approx - 0.75317 \dots, u \approx 1.07169 \dots, u \approx - 1.65186 \dots

8 (1 - u^{2}) - 6 u^{3} + 8 u = 0

展开

8 (1 - u^{2}) - 6 u^{3} + 8 u : 8 - 8 u^{2} - 6 u^{3} + 8 u

8 (1 - u^{2}) - 6 u^{3} + 8 u

乘开

8 (1 - u^{2}) : 8 - 8 u^{2}

8 (1 - u^{2})

使用分配律:

a (b - c) = ab - a c

a = 8, b = 1, c = u^{2}

= 8 \cdot 1 - 8 u^{2}

数字相乘:

8 \cdot 1 = 8

= 8 - 8 u^{2}

= 8 - 8 u^{2} - 6 u^{3} + 8 u

8 - 8 u^{2} - 6 u^{3} + 8 u = 0

改写成标准形式

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

- 6 u^{3} - 8 u^{2} + 8 u + 8 = 0

使用牛顿-拉弗森方法找到

- 6 u^{3} - 8 u^{2} + 8 u + 8 = 0

的一个解:

u \approx - 0.75317 \dots

- 6 u^{3} - 8 u^{2} + 8 u + 8 = 0

牛顿-拉弗森近似法定义

f (u) = - 6 u^{3} - 8 u^{2} + 8 u + 8

找到

f^{^{'}} (u) : - 18 u^{2} - 16 u + 8

\frac{d}{d u} (- 6 u^{3} - 8 u^{2} + 8 u + 8)

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= - \frac{d}{d u} (6 u^{3}) - \frac{d}{d u} (8 u^{2}) + \frac{d}{d u} (8 u) + \frac{d}{d u} (8)

\frac{d}{d u} (6 u^{3}) = 18 u^{2}

\frac{d}{d u} (6 u^{3})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 6 \frac{d}{d u} (u^{3})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 6 \cdot 3 u^{3 - 1}

化简

= 18 u^{2}

\frac{d}{d u} (8 u^{2}) = 16 u

\frac{d}{d u} (8 u^{2})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 8 \frac{d}{d u} (u^{2})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 8 \cdot 2 u^{2 - 1}

化简

= 16 u

\frac{d}{d u} (8 u) = 8

\frac{d}{d u} (8 u)

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 8 \frac{d u}{d u}

使用常见微分定则:

\frac{d u}{d u} = 1

= 8 \cdot 1

化简

= 8

\frac{d}{d u} (8) = 0

\frac{d}{d u} (8)

常数微分:

\frac{d}{d x} (a) = 0

= 0

= - 18 u^{2} - 16 u + 8 + 0

化简

= - 18 u^{2} - 16 u + 8

令

u_{0} = - 1

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = - 0.66666 \dots : Δ u_{1} = 0.33333 \dots

f (u_{0}) = - 6 (- 1)^{3} - 8 (- 1)^{2} + 8 (- 1) + 8 = - 2

f^{^{'}} (u_{0}) = - 18 (- 1)^{2} - 16 (- 1) + 8 = 6

u_{1} = - 0.66666 \dots

Δ u_{1} = ∣ - 0.66666 \dots - (- 1) ∣ = 0.33333 \dots

Δ u_{1} = 0.33333 \dots

u_{2} = - 0.75 : Δ u_{2} = 0.08333 \dots

f (u_{1}) = - 6 (- 0.66666 \dots)^{3} - 8 (- 0.66666 \dots)^{2} + 8 (- 0.66666 \dots) + 8 = 0.88888 \dots

f^{^{'}} (u_{1}) = - 18 (- 0.66666 \dots)^{2} - 16 (- 0.66666 \dots) + 8 = 10.66666 \dots

u_{2} = - 0.75

Δ u_{2} = ∣ - 0.75 - (- 0.66666 \dots) ∣ = 0.08333 \dots

Δ u_{2} = 0.08333 \dots

u_{3} = - 0.75316 \dots : Δ u_{3} = 0.00316 \dots

f (u_{2}) = - 6 (- 0.75)^{3} - 8 (- 0.75)^{2} + 8 (- 0.75) + 8 = 0.03125

f^{^{'}} (u_{2}) = - 18 (- 0.75)^{2} - 16 (- 0.75) + 8 = 9.875

u_{3} = - 0.75316 \dots

Δ u_{3} = ∣ - 0.75316 \dots - (- 0.75) ∣ = 0.00316 \dots

Δ u_{3} = 0.00316 \dots

u_{4} = - 0.75317 \dots : Δ u_{4} = 5.61681 E - 6

f (u_{3}) = - 6 (- 0.75316 \dots)^{3} - 8 (- 0.75316 \dots)^{2} + 8 (- 0.75316 \dots) + 8 = 0.00005 \dots

f^{^{'}} (u_{3}) = - 18 (- 0.75316 \dots)^{2} - 16 (- 0.75316 \dots) + 8 = 9.84000 \dots

u_{4} = - 0.75317 \dots

Δ u_{4} = ∣ - 0.75317 \dots - (- 0.75316 \dots) ∣ = 5.61681 E - 6

Δ u_{4} = 5.61681 E - 6

u_{5} = - 0.75317 \dots : Δ u_{5} = 1.78166 E - 11

f (u_{4}) = - 6 (- 0.75317 \dots)^{3} - 8 (- 0.75317 \dots)^{2} + 8 (- 0.75317 \dots) + 8 = 1.75314 E - 10

f^{^{'}} (u_{4}) = - 18 (- 0.75317 \dots)^{2} - 16 (- 0.75317 \dots) + 8 = 9.83994 \dots

u_{5} = - 0.75317 \dots

Δ u_{5} = ∣ - 0.75317 \dots - (- 0.75317 \dots) ∣ = 1.78166 E - 11

Δ u_{5} = 1.78166 E - 11

u \approx - 0.75317 \dots

使用长除法

Eq u a t i o n 0 : \frac{- 6 u ^{3} - 8 u ^{2} + 8 u + 8}{u + 0.75317 \dots} = - 6 u^{2} - 3.48097 \dots u + 10.62176 \dots

- 6 u^{2} - 3.48097 \dots u + 10.62176 \dots \approx 0

使用牛顿-拉弗森方法找到

- 6 u^{2} - 3.48097 \dots u + 10.62176 \dots = 0

的一个解:

u \approx 1.07169 \dots

- 6 u^{2} - 3.48097 \dots u + 10.62176 \dots = 0

牛顿-拉弗森近似法定义

f (u) = - 6 u^{2} - 3.48097 \dots u + 10.62176 \dots

找到

f^{^{'}} (u) : - 12 u - 3.48097 \dots

\frac{d}{d u} (- 6 u^{2} - 3.48097 \dots u + 10.62176 \dots)

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= - \frac{d}{d u} (6 u^{2}) - \frac{d}{d u} (3.48097 \dots u) + \frac{d}{d u} (10.62176 \dots)

\frac{d}{d u} (6 u^{2}) = 12 u

\frac{d}{d u} (6 u^{2})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 6 \frac{d}{d u} (u^{2})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 6 \cdot 2 u^{2 - 1}

化简

= 12 u

\frac{d}{d u} (3.48097 \dots u) = 3.48097 \dots

\frac{d}{d u} (3.48097 \dots u)

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 3.48097 \dots \frac{d u}{d u}

使用常见微分定则:

\frac{d u}{d u} = 1

= 3.48097 \dots \cdot 1

化简

= 3.48097 \dots

\frac{d}{d u} (10.62176 \dots) = 0

\frac{d}{d u} (10.62176 \dots)

常数微分:

\frac{d}{d x} (a) = 0

= 0

= - 12 u - 3.48097 \dots + 0

化简

= - 12 u - 3.48097 \dots

令

u_{0} = 3

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 1.63678 \dots : Δ u_{1} = 1.36321 \dots

f (u_{0}) = - 6 \cdot 3^{2} - 3.48097 \dots \cdot 3 + 10.62176 \dots = - 53.82116 \dots

f^{^{'}} (u_{0}) = - 12 \cdot 3 - 3.48097 \dots = - 39.48097 \dots

u_{1} = 1.63678 \dots

Δ u_{1} = ∣ 1.63678 \dots - 3 ∣ = 1.36321 \dots

Δ u_{1} = 1.36321 \dots

u_{2} = 1.15455 \dots : Δ u_{2} = 0.48222 \dots

f (u_{1}) = - 6 \cdot 1.63678 \dots^{2} - 3.48097 \dots \cdot 1.63678 \dots + 10.62176 \dots = - 11.15017 \dots

f^{^{'}} (u_{1}) = - 12 \cdot 1.63678 \dots - 3.48097 \dots = - 23.12236 \dots

u_{2} = 1.15455 \dots

Δ u_{2} = ∣ 1.15455 \dots - 1.63678 \dots ∣ = 0.48222 \dots

Δ u_{2} = 0.48222 \dots

u_{3} = 1.07407 \dots : Δ u_{3} = 0.08048 \dots

f (u_{2}) = - 6 \cdot 1.15455 \dots^{2} - 3.48097 \dots \cdot 1.15455 \dots + 10.62176 \dots = - 1.39524 \dots

f^{^{'}} (u_{2}) = - 12 \cdot 1.15455 \dots - 3.48097 \dots = - 17.33567 \dots

u_{3} = 1.07407 \dots

Δ u_{3} = ∣ 1.07407 \dots - 1.15455 \dots ∣ = 0.08048 \dots

Δ u_{3} = 0.08048 \dots

u_{4} = 1.07169 \dots : Δ u_{4} = 0.00237 \dots

f (u_{3}) = - 6 \cdot 1.07407 \dots^{2} - 3.48097 \dots \cdot 1.07407 \dots + 10.62176 \dots = - 0.03886 \dots

f^{^{'}} (u_{3}) = - 12 \cdot 1.07407 \dots - 3.48097 \dots = - 16.36986 \dots

u_{4} = 1.07169 \dots

Δ u_{4} = ∣ 1.07169 \dots - 1.07407 \dots ∣ = 0.00237 \dots

Δ u_{4} = 0.00237 \dots

u_{5} = 1.07169 \dots : Δ u_{5} = 2.06972 E - 6

f (u_{4}) = - 6 \cdot 1.07169 \dots^{2} - 3.48097 \dots \cdot 1.07169 \dots + 10.62176 \dots = - 0.00003 \dots

f^{^{'}} (u_{4}) = - 12 \cdot 1.07169 \dots - 3.48097 \dots = - 16.34137 \dots

u_{5} = 1.07169 \dots

Δ u_{5} = ∣ 1.07169 \dots - 1.07169 \dots ∣ = 2.06972 E - 6

Δ u_{5} = 2.06972 E - 6

u_{6} = 1.07169 \dots : Δ u_{6} = 1.57283 E - 12

f (u_{5}) = - 6 \cdot 1.07169 \dots^{2} - 3.48097 \dots \cdot 1.07169 \dots + 10.62176 \dots = - 2.57021 E - 11

f^{^{'}} (u_{5}) = - 12 \cdot 1.07169 \dots - 3.48097 \dots = - 16.34134 \dots

u_{6} = 1.07169 \dots

Δ u_{6} = ∣ 1.07169 \dots - 1.07169 \dots ∣ = 1.57283 E - 12

Δ u_{6} = 1.57283 E - 12

u \approx 1.07169 \dots

使用长除法

Eq u a t i o n 0 : \frac{- 6 u ^{2} - 3.48097 \dots u + 10.62176 \dots}{u - 1.07169 \dots} = - 6 u - 9.91116 \dots

- 6 u - 9.91116 \dots \approx 0

u \approx - 1.65186 \dots

解为

u \approx - 0.75317 \dots, u \approx 1.07169 \dots, u \approx - 1.65186 \dots

u \approx - 0.75317 \dots, u \approx 1.07169 \dots, u \approx - 1.65186 \dots

验证解

找到无定义的点（奇点）:

u = 1, u = - 1

取

8 + \frac{- 6 u ^{3} + 8 u}{1 - u ^{2}}

的分母，令其等于零

解

1 - u^{2} = 0 : u = 1, u = - 1

1 - u^{2} = 0

将

1

到右边

1 - u^{2} = 0

两边减去

1

1 - u^{2} - 1 = 0 - 1

化简

- u^{2} = - 1

- u^{2} = - 1

两边除以

- 1

- u^{2} = - 1

两边除以

- 1

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

化简

u^{2} = 1

u^{2} = 1

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

使用根式运算法则:

1 = 1

= 1

- 1 = - 1

- 1

使用根式运算法则:

1 = 1

1 = 1

= - 1

u = 1, u = - 1

以下点无定义

u = 1, u = - 1

将不在定义域的点与解相综合:

u \approx - 0.75317 \dots, u \approx 1.07169 \dots, u \approx - 1.65186 \dots

u = tan (θ)

代回

tan (θ) \approx - 0.75317 \dots, tan (θ) \approx 1.07169 \dots, tan (θ) \approx - 1.65186 \dots

tan (θ) \approx - 0.75317 \dots, tan (θ) \approx 1.07169 \dots, tan (θ) \approx - 1.65186 \dots

tan (θ) = - 0.75317 \dots : θ = arctan (- 0.75317 \dots) + πn

tan (θ) = - 0.75317 \dots

使用反三角函数性质

tan (θ) = - 0.75317 \dots

tan (θ) = - 0.75317 \dots

的通解

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

θ = arctan (- 0.75317 \dots) + πn

θ = arctan (- 0.75317 \dots) + πn

tan (θ) = 1.07169 \dots : θ = arctan (1.07169 \dots) + πn

tan (θ) = 1.07169 \dots

使用反三角函数性质

tan (θ) = 1.07169 \dots

tan (θ) = 1.07169 \dots

的通解

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

θ = arctan (1.07169 \dots) + πn

θ = arctan (1.07169 \dots) + πn

tan (θ) = - 1.65186 \dots : θ = arctan (- 1.65186 \dots) + πn

tan (θ) = - 1.65186 \dots

使用反三角函数性质

tan (θ) = - 1.65186 \dots

tan (θ) = - 1.65186 \dots

的通解

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

θ = arctan (- 1.65186 \dots) + πn

θ = arctan (- 1.65186 \dots) + πn

合并所有解

θ = arctan (- 0.75317 \dots) + πn, θ = arctan (1.07169 \dots) + πn, θ = arctan (- 1.65186 \dots) + πn

以小数形式表示解

θ = - 0.64552 \dots + πn, θ = 0.81999 \dots + πn, θ = - 1.02643 \dots + πn

作图

流行的例子

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\frac{tan ^{2} ( x )}{sec ( x ) + 1} = tan (x)

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