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受欢迎的三角函数 >

csc^2(x)-3=6tan(x)

解答

csc^{2} (x) - 3 = 6 tan (x)

解答

x = 0.43008 \dots + πn

+1

度数

x = 24.64185 \dots^{\circ} + 18 0^{\circ} n

求解步骤

csc^{2} (x) - 3 = 6 tan (x)

两边进行平方

(csc^{2} (x) - 3)^{2} = (6 tan (x))^{2}

两边减去

(6 tan (x))^{2}

(csc^{2} (x) - 3)^{2} - 36 tan^{2} (x) = 0

使用三角恒等式改写

(- 3 + csc^{2} (x))^{2} - 36 tan^{2} (x)

使用毕达哥拉斯恒等式:

csc^{2} (x) = 1 + cot^{2} (x)

= (- 3 + 1 + cot^{2} (x))^{2} - 36 tan^{2} (x)

化简

= (cot^{2} (x) - 2)^{2} - 36 tan^{2} (x)

(- 2 + cot^{2} (x))^{2} - 36 tan^{2} (x) = 0

分解

(- 2 + cot^{2} (x))^{2} - 36 tan^{2} (x) : (- 2 + cot^{2} (x) + 6 tan (x)) (- 2 + cot^{2} (x) - 6 tan (x))

(- 2 + cot^{2} (x))^{2} - 36 tan^{2} (x)

将

(- 2 + cot^{2} (x))^{2} - 36 tan^{2} (x)

改写为

(- 2 + cot^{2} (x))^{2} - (6 tan (x))^{2}

(- 2 + cot^{2} (x))^{2} - 36 tan^{2} (x)

将

36

改写为

6^{2}

= (- 2 + cot^{2} (x))^{2} - 6^{2} tan^{2} (x)

使用指数法则:

a^{m} b^{m} = (ab)^{m}

6^{2} tan^{2} (x) = (6 tan (x))^{2}

= (- 2 + cot^{2} (x))^{2} - (6 tan (x))^{2}

= (- 2 + cot^{2} (x))^{2} - (6 tan (x))^{2}

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

(- 2 + cot^{2} (x))^{2} - (6 tan (x))^{2} = ((- 2 + cot^{2} (x)) + 6 tan (x)) ((- 2 + cot^{2} (x)) - 6 tan (x))

= ((- 2 + cot^{2} (x)) + 6 tan (x)) ((- 2 + cot^{2} (x)) - 6 tan (x))

整理后得

= (cot^{2} (x) + 6 tan (x) - 2) (cot^{2} (x) - 6 tan (x) - 2)

(- 2 + cot^{2} (x) + 6 tan (x)) (- 2 + cot^{2} (x) - 6 tan (x)) = 0

分别求解每个部分

- 2 + cot^{2} (x) + 6 tan (x) = 0 or - 2 + cot^{2} (x) - 6 tan (x) = 0

- 2 + cot^{2} (x) + 6 tan (x) = 0 : x = arccot (- 2.17998 \dots) + πn

- 2 + cot^{2} (x) + 6 tan (x) = 0

使用三角恒等式改写

- 2 + cot^{2} (x) + 6 tan (x)

使用基本三角恒等式:

tan (x) = \frac{1}{cot ( x )}

= - 2 + cot^{2} (x) + 6 \cdot \frac{1}{cot ( x )}

6 \cdot \frac{1}{cot ( x )} = \frac{6}{cot ( x )}

6 \cdot \frac{1}{cot ( x )}

分式相乘:

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

= \frac{1 \cdot 6}{cot ( x )}

数字相乘:

1 \cdot 6 = 6

= \frac{6}{cot ( x )}

= - 2 + cot^{2} (x) + \frac{6}{cot ( x )}

- 2 + cot^{2} (x) + \frac{6}{cot ( x )} = 0

用替代法求解

- 2 + cot^{2} (x) + \frac{6}{cot ( x )} = 0

令

: cot (x) = u

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

- 2 + u^{2} + \frac{6}{u} = 0 : u \approx - 2.17998 \dots

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

在两边乘以

u

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

在两边乘以

u

- 2 u + u^{2} u + \frac{6}{u} u = 0 \cdot u

化简

- 2 u + u^{2} u + \frac{6}{u} u = 0 \cdot u

化简

u^{2} u : u^{3}

u^{2} u

使用指数法则:

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

u^{2} u = u^{2 + 1}

= u^{2 + 1}

数字相加:

2 + 1 = 3

= u^{3}

化简

\frac{6}{u} u : 6

\frac{6}{u} u

分式相乘:

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

= \frac{6 u}{u}

约分:

u

= 6

化简

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

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

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

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

解

- 2 u + u^{3} + 6 = 0 : u \approx - 2.17998 \dots

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

改写成标准形式

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

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

使用牛顿-拉弗森方法找到

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

的一个解:

u \approx - 2.17998 \dots

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

牛顿-拉弗森近似法定义

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

找到

f^{^{'}} (u) : 3 u^{2} - 2

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

使用微分加减法定则:

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

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

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

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

使用幂法则:

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

= 3 u^{3 - 1}

化简

= 3 u^{2}

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

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

将常数提出:

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

= 2 \frac{d u}{d u}

使用常见微分定则:

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

= 2 \cdot 1

化简

= 2

\frac{d}{d u} (6) = 0

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

常数微分:

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

= 0

= 3 u^{2} - 2 + 0

化简

= 3 u^{2} - 2

令

u_{0} = 3

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 1.92 : Δ u_{1} = 1.08

f (u_{0}) = 3^{3} - 2 \cdot 3 + 6 = 27

f^{^{'}} (u_{0}) = 3 \cdot 3^{2} - 2 = 25

u_{1} = 1.92

Δ u_{1} = ∣ 1.92 - 3 ∣ = 1.08

Δ u_{1} = 1.08

u_{2} = 0.90027 \dots : Δ u_{2} = 1.01972 \dots

f (u_{1}) = 1.9 2^{3} - 2 \cdot 1.92 + 6 = 9.237888

f^{^{'}} (u_{1}) = 3 \cdot 1.9 2^{2} - 2 = 9.0592

u_{2} = 0.90027 \dots

Δ u_{2} = ∣ 0.90027 \dots - 1.92 ∣ = 1.01972 \dots

Δ u_{2} = 1.01972 \dots

u_{3} = - 10.52325 \dots : Δ u_{3} = 11.42353 \dots

f (u_{2}) = 0.90027 \dots^{3} - 2 \cdot 0.90027 \dots + 6 = 4.92911 \dots

f^{^{'}} (u_{2}) = 3 \cdot 0.90027 \dots^{2} - 2 = 0.43148 \dots

u_{3} = - 10.52325 \dots

Δ u_{3} = ∣ - 10.52325 \dots - 0.90027 \dots ∣ = 11.42353 \dots

Δ u_{3} = 11.42353 \dots

u_{4} = - 7.07616 \dots : Δ u_{4} = 3.44709 \dots

f (u_{3}) = (- 10.52325 \dots)^{3} - 2 (- 10.52325 \dots) + 6 = - 1138.28856 \dots

f^{^{'}} (u_{3}) = 3 (- 10.52325 \dots)^{2} - 2 = 330.21696 \dots

u_{4} = - 7.07616 \dots

Δ u_{4} = ∣ - 7.07616 \dots - (- 10.52325 \dots) ∣ = 3.44709 \dots

Δ u_{4} = 3.44709 \dots

u_{5} = - 4.82158 \dots : Δ u_{5} = 2.25458 \dots

f (u_{4}) = (- 7.07616 \dots)^{3} - 2 (- 7.07616 \dots) + 6 = - 334.16639 \dots

f^{^{'}} (u_{4}) = 3 (- 7.07616 \dots)^{2} - 2 = 148.21639 \dots

u_{5} = - 4.82158 \dots

Δ u_{5} = ∣ - 4.82158 \dots - (- 7.07616 \dots) ∣ = 2.25458 \dots

Δ u_{5} = 2.25458 \dots

u_{6} = - 3.39785 \dots : Δ u_{6} = 1.42372 \dots

f (u_{5}) = (- 4.82158 \dots)^{3} - 2 (- 4.82158 \dots) + 6 = - 96.44728 \dots

f^{^{'}} (u_{5}) = 3 (- 4.82158 \dots)^{2} - 2 = 67.74295 \dots

u_{6} = - 3.39785 \dots

Δ u_{6} = ∣ - 3.39785 \dots - (- 4.82158 \dots) ∣ = 1.42372 \dots

Δ u_{6} = 1.42372 \dots

u_{7} = - 2.58789 \dots : Δ u_{7} = 0.80995 \dots

f (u_{6}) = (- 3.39785 \dots)^{3} - 2 (- 3.39785 \dots) + 6 = - 26.43402 \dots

f^{^{'}} (u_{6}) = 3 (- 3.39785 \dots)^{2} - 2 = 32.63630 \dots

u_{7} = - 2.58789 \dots

Δ u_{7} = ∣ - 2.58789 \dots - (- 3.39785 \dots) ∣ = 0.80995 \dots

Δ u_{7} = 0.80995 \dots

u_{8} = - 2.24763 \dots : Δ u_{8} = 0.34026 \dots

f (u_{7}) = (- 2.58789 \dots)^{3} - 2 (- 2.58789 \dots) + 6 = - 6.15594 \dots

f^{^{'}} (u_{7}) = 3 (- 2.58789 \dots)^{2} - 2 = 18.09167 \dots

u_{8} = - 2.24763 \dots

Δ u_{8} = ∣ - 2.24763 \dots - (- 2.58789 \dots) ∣ = 0.34026 \dots

Δ u_{8} = 0.34026 \dots

u_{9} = - 2.18230 \dots : Δ u_{9} = 0.06533 \dots

f (u_{8}) = (- 2.24763 \dots)^{3} - 2 (- 2.24763 \dots) + 6 = - 0.85948 \dots

f^{^{'}} (u_{8}) = 3 (- 2.24763 \dots)^{2} - 2 = 13.15559 \dots

u_{9} = - 2.18230 \dots

Δ u_{9} = ∣ - 2.18230 \dots - (- 2.24763 \dots) ∣ = 0.06533 \dots

Δ u_{9} = 0.06533 \dots

u_{10} = - 2.17998 \dots : Δ u_{10} = 0.00231 \dots

f (u_{9}) = (- 2.18230 \dots)^{3} - 2 (- 2.18230 \dots) + 6 = - 0.02850 \dots

f^{^{'}} (u_{9}) = 3 (- 2.18230 \dots)^{2} - 2 = 12.28734 \dots

u_{10} = - 2.17998 \dots

Δ u_{10} = ∣ - 2.17998 \dots - (- 2.18230 \dots) ∣ = 0.00231 \dots

Δ u_{10} = 0.00231 \dots

u_{11} = - 2.17998 \dots : Δ u_{11} = 2.87294 E - 6

f (u_{10}) = (- 2.17998 \dots)^{3} - 2 (- 2.17998 \dots) + 6 = - 0.00003 \dots

f^{^{'}} (u_{10}) = 3 (- 2.17998 \dots)^{2} - 2 = 12.25699 \dots

u_{11} = - 2.17998 \dots

Δ u_{11} = ∣ - 2.17998 \dots - (- 2.17998 \dots) ∣ = 2.87294 E - 6

Δ u_{11} = 2.87294 E - 6

u_{12} = - 2.17998 \dots : Δ u_{12} = 4.4041 E - 12

f (u_{11}) = (- 2.17998 \dots)^{3} - 2 (- 2.17998 \dots) + 6 = - 5.39808 E - 11

f^{^{'}} (u_{11}) = 3 (- 2.17998 \dots)^{2} - 2 = 12.25695 \dots

u_{12} = - 2.17998 \dots

Δ u_{12} = ∣ - 2.17998 \dots - (- 2.17998 \dots) ∣ = 4.4041 E - 12

Δ u_{12} = 4.4041 E - 12

u \approx - 2.17998 \dots

使用长除法

Eq u a t i o n 0 : \frac{u ^{3} - 2 u + 6}{u + 2.17998 \dots} = u^{2} - 2.17998 \dots u + 2.75231 \dots

u^{2} - 2.17998 \dots u + 2.75231 \dots \approx 0

使用牛顿-拉弗森方法找到

u^{2} - 2.17998 \dots u + 2.75231 \dots = 0

的一个解:

u \in R

无解

u^{2} - 2.17998 \dots u + 2.75231 \dots = 0

牛顿-拉弗森近似法定义

f (u) = u^{2} - 2.17998 \dots u + 2.75231 \dots

找到

f^{^{'}} (u) : 2 u - 2.17998 \dots

\frac{d}{d u} (u^{2} - 2.17998 \dots u + 2.75231 \dots)

使用微分加减法定则:

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

= \frac{d}{d u} (u^{2}) - \frac{d}{d u} (2.17998 \dots u) + \frac{d}{d u} (2.75231 \dots)

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

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

使用幂法则:

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

= 2 u^{2 - 1}

化简

= 2 u

\frac{d}{d u} (2.17998 \dots u) = 2.17998 \dots

\frac{d}{d u} (2.17998 \dots u)

将常数提出:

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

= 2.17998 \dots \frac{d u}{d u}

使用常见微分定则:

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

= 2.17998 \dots \cdot 1

化简

= 2.17998 \dots

\frac{d}{d u} (2.75231 \dots) = 0

\frac{d}{d u} (2.75231 \dots)

常数微分:

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

= 0

= 2 u - 2.17998 \dots + 0

化简

= 2 u - 2.17998 \dots

令

u_{0} = 1

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 9.73612 \dots : Δ u_{1} = 8.73612 \dots

f (u_{0}) = 1^{2} - 2.17998 \dots \cdot 1 + 2.75231 \dots = 1.57233 \dots

f^{^{'}} (u_{0}) = 2 \cdot 1 - 2.17998 \dots = - 0.17998 \dots

u_{1} = 9.73612 \dots

Δ u_{1} = ∣ 9.73612 \dots - 1 ∣ = 8.73612 \dots

Δ u_{1} = 8.73612 \dots

u_{2} = 5.32259 \dots : Δ u_{2} = 4.41352 \dots

f (u_{1}) = 9.73612 \dots^{2} - 2.17998 \dots \cdot 9.73612 \dots + 2.75231 \dots = 76.31980 \dots

f^{^{'}} (u_{1}) = 2 \cdot 9.73612 \dots - 2.17998 \dots = 17.29226 \dots

u_{2} = 5.32259 \dots

Δ u_{2} = ∣ 5.32259 \dots - 9.73612 \dots ∣ = 4.41352 \dots

Δ u_{2} = 4.41352 \dots

u_{3} = 3.02150 \dots : Δ u_{3} = 2.30108 \dots

f (u_{2}) = 5.32259 \dots^{2} - 2.17998 \dots \cdot 5.32259 \dots + 2.75231 \dots = 19.47919 \dots

f^{^{'}} (u_{2}) = 2 \cdot 5.32259 \dots - 2.17998 \dots = 8.46521 \dots

u_{3} = 3.02150 \dots

Δ u_{3} = ∣ 3.02150 \dots - 5.32259 \dots ∣ = 2.30108 \dots

Δ u_{3} = 2.30108 \dots

u_{4} = 1.65082 \dots : Δ u_{4} = 1.37068 \dots

f (u_{3}) = 3.02150 \dots^{2} - 2.17998 \dots \cdot 3.02150 \dots + 2.75231 \dots = 5.29500 \dots

f^{^{'}} (u_{3}) = 2 \cdot 3.02150 \dots - 2.17998 \dots = 3.86303 \dots

u_{4} = 1.65082 \dots

Δ u_{4} = ∣ 1.65082 \dots - 3.02150 \dots ∣ = 1.37068 \dots

Δ u_{4} = 1.37068 \dots

u_{5} = - 0.02415 \dots : Δ u_{5} = 1.67497 \dots

f (u_{4}) = 1.65082 \dots^{2} - 2.17998 \dots \cdot 1.65082 \dots + 2.75231 \dots = 1.87877 \dots

f^{^{'}} (u_{4}) = 2 \cdot 1.65082 \dots - 2.17998 \dots = 1.12167 \dots

u_{5} = - 0.02415 \dots

Δ u_{5} = ∣ - 0.02415 \dots - 1.65082 \dots ∣ = 1.67497 \dots

Δ u_{5} = 1.67497 \dots

u_{6} = 1.23490 \dots : Δ u_{6} = 1.25906 \dots

f (u_{5}) = (- 0.02415 \dots)^{2} - 2.17998 \dots (- 0.02415 \dots) + 2.75231 \dots = 2.80555 \dots

f^{^{'}} (u_{5}) = 2 (- 0.02415 \dots) - 2.17998 \dots = - 2.22828 \dots

u_{6} = 1.23490 \dots

Δ u_{6} = ∣ 1.23490 \dots - (- 0.02415 \dots) ∣ = 1.25906 \dots

Δ u_{6} = 1.25906 \dots

u_{7} = - 4.23449 \dots : Δ u_{7} = 5.46939 \dots

f (u_{6}) = 1.23490 \dots^{2} - 2.17998 \dots \cdot 1.23490 \dots + 2.75231 \dots = 1.58523 \dots

f^{^{'}} (u_{6}) = 2 \cdot 1.23490 \dots - 2.17998 \dots = 0.28983 \dots

u_{7} = - 4.23449 \dots

Δ u_{7} = ∣ - 4.23449 \dots - 1.23490 \dots ∣ = 5.46939 \dots

Δ u_{7} = 5.46939 \dots

u_{8} = - 1.42535 \dots : Δ u_{8} = 2.80913 \dots

f (u_{7}) = (- 4.23449 \dots)^{2} - 2.17998 \dots (- 4.23449 \dots) + 2.75231 \dots = 29.91433 \dots

f^{^{'}} (u_{7}) = 2 (- 4.23449 \dots) - 2.17998 \dots = - 10.64896 \dots

u_{8} = - 1.42535 \dots

Δ u_{8} = ∣ - 1.42535 \dots - (- 4.23449 \dots) ∣ = 2.80913 \dots

Δ u_{8} = 2.80913 \dots

u_{9} = 0.14325 \dots : Δ u_{9} = 1.56861 \dots

f (u_{8}) = (- 1.42535 \dots)^{2} - 2.17998 \dots (- 1.42535 \dots) + 2.75231 \dots = 7.89122 \dots

f^{^{'}} (u_{8}) = 2 (- 1.42535 \dots) - 2.17998 \dots = - 5.03069 \dots

u_{9} = 0.14325 \dots

Δ u_{9} = ∣ 0.14325 \dots - (- 1.42535 \dots) ∣ = 1.56861 \dots

Δ u_{9} = 1.56861 \dots

u_{10} = 1.44274 \dots : Δ u_{10} = 1.29948 \dots

f (u_{9}) = 0.14325 \dots^{2} - 2.17998 \dots \cdot 0.14325 \dots + 2.75231 \dots = 2.46054 \dots

f^{^{'}} (u_{9}) = 2 \cdot 0.14325 \dots - 2.17998 \dots = - 1.89347 \dots

u_{10} = 1.44274 \dots

Δ u_{10} = ∣ 1.44274 \dots - 0.14325 \dots ∣ = 1.29948 \dots

Δ u_{10} = 1.29948 \dots

u_{11} = - 0.95081 \dots : Δ u_{11} = 2.39356 \dots

f (u_{10}) = 1.44274 \dots^{2} - 2.17998 \dots \cdot 1.44274 \dots + 2.75231 \dots = 1.68867 \dots

f^{^{'}} (u_{10}) = 2 \cdot 1.44274 \dots - 2.17998 \dots = 0.70550 \dots

u_{11} = - 0.95081 \dots

Δ u_{11} = ∣ - 0.95081 \dots - 1.44274 \dots ∣ = 2.39356 \dots

Δ u_{11} = 2.39356 \dots

u_{12} = 0.45282 \dots : Δ u_{12} = 1.40364 \dots

f (u_{11}) = (- 0.95081 \dots)^{2} - 2.17998 \dots (- 0.95081 \dots) + 2.75231 \dots = 5.72913 \dots

f^{^{'}} (u_{11}) = 2 (- 0.95081 \dots) - 2.17998 \dots = - 4.08161 \dots

u_{12} = 0.45282 \dots

Δ u_{12} = ∣ 0.45282 \dots - (- 0.95081 \dots) ∣ = 1.40364 \dots

Δ u_{12} = 1.40364 \dots

u_{13} = 1.99891 \dots : Δ u_{13} = 1.54608 \dots

f (u_{12}) = 0.45282 \dots^{2} - 2.17998 \dots \cdot 0.45282 \dots + 2.75231 \dots = 1.97021 \dots

f^{^{'}} (u_{12}) = 2 \cdot 0.45282 \dots - 2.17998 \dots = - 1.27432 \dots

u_{13} = 1.99891 \dots

Δ u_{13} = ∣ 1.99891 \dots - 0.45282 \dots ∣ = 1.54608 \dots

Δ u_{13} = 1.54608 \dots

无法得出解

解是

u \approx - 2.17998 \dots

u \approx - 2.17998 \dots

验证解

找到无定义的点（奇点）:

u = 0

取

- 2 + u^{2} + \frac{6}{u}

的分母，令其等于零

u = 0

以下点无定义

u = 0

将不在定义域的点与解相综合:

u \approx - 2.17998 \dots

u = cot (x)

代回

cot (x) \approx - 2.17998 \dots

cot (x) \approx - 2.17998 \dots

cot (x) = - 2.17998 \dots : x = arccot (- 2.17998 \dots) + πn

cot (x) = - 2.17998 \dots

使用反三角函数性质

cot (x) = - 2.17998 \dots

cot (x) = - 2.17998 \dots

的通解

cot (x) = - a \Rightarrow x = arccot (- a) + πn

x = arccot (- 2.17998 \dots) + πn

x = arccot (- 2.17998 \dots) + πn

合并所有解

x = arccot (- 2.17998 \dots) + πn

- 2 + cot^{2} (x) - 6 tan (x) = 0 : x = arccot (2.17998 \dots) + πn

- 2 + cot^{2} (x) - 6 tan (x) = 0

使用三角恒等式改写

- 2 + cot^{2} (x) - 6 tan (x)

使用基本三角恒等式:

tan (x) = \frac{1}{cot ( x )}

= - 2 + cot^{2} (x) - 6 \cdot \frac{1}{cot ( x )}

6 \cdot \frac{1}{cot ( x )} = \frac{6}{cot ( x )}

6 \cdot \frac{1}{cot ( x )}

分式相乘:

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

= \frac{1 \cdot 6}{cot ( x )}

数字相乘:

1 \cdot 6 = 6

= \frac{6}{cot ( x )}

= - 2 + cot^{2} (x) - \frac{6}{cot ( x )}

- 2 + cot^{2} (x) - \frac{6}{cot ( x )} = 0

用替代法求解

- 2 + cot^{2} (x) - \frac{6}{cot ( x )} = 0

令

: cot (x) = u

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

- 2 + u^{2} - \frac{6}{u} = 0 : u \approx 2.17998 \dots

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

在两边乘以

u

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

在两边乘以

u

- 2 u + u^{2} u - \frac{6}{u} u = 0 \cdot u

化简

- 2 u + u^{2} u - \frac{6}{u} u = 0 \cdot u

化简

u^{2} u : u^{3}

u^{2} u

使用指数法则:

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

u^{2} u = u^{2 + 1}

= u^{2 + 1}

数字相加:

2 + 1 = 3

= u^{3}

化简

- \frac{6}{u} u : - 6

- \frac{6}{u} u

分式相乘:

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

= - \frac{6 u}{u}

约分:

u

= - 6

化简

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

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

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

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

解

- 2 u + u^{3} - 6 = 0 : u \approx 2.17998 \dots

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

改写成标准形式

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

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

使用牛顿-拉弗森方法找到

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

的一个解:

u \approx 2.17998 \dots

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

牛顿-拉弗森近似法定义

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

找到

f^{^{'}} (u) : 3 u^{2} - 2

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

使用微分加减法定则:

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

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

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

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

使用幂法则:

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

= 3 u^{3 - 1}

化简

= 3 u^{2}

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

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

将常数提出:

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

= 2 \frac{d u}{d u}

使用常见微分定则:

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

= 2 \cdot 1

化简

= 2

\frac{d}{d u} (6) = 0

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

常数微分:

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

= 0

= 3 u^{2} - 2 - 0

化简

= 3 u^{2} - 2

令

u_{0} = - 3

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = - 1.92 : Δ u_{1} = 1.08

f (u_{0}) = (- 3)^{3} - 2 (- 3) - 6 = - 27

f^{^{'}} (u_{0}) = 3 (- 3)^{2} - 2 = 25

u_{1} = - 1.92

Δ u_{1} = ∣ - 1.92 - (- 3) ∣ = 1.08

Δ u_{1} = 1.08

u_{2} = - 0.90027 \dots : Δ u_{2} = 1.01972 \dots

f (u_{1}) = (- 1.92)^{3} - 2 (- 1.92) - 6 = - 9.237888

f^{^{'}} (u_{1}) = 3 (- 1.92)^{2} - 2 = 9.0592

u_{2} = - 0.90027 \dots

Δ u_{2} = ∣ - 0.90027 \dots - (- 1.92) ∣ = 1.01972 \dots

Δ u_{2} = 1.01972 \dots

u_{3} = 10.52325 \dots : Δ u_{3} = 11.42353 \dots

f (u_{2}) = (- 0.90027 \dots)^{3} - 2 (- 0.90027 \dots) - 6 = - 4.92911 \dots

f^{^{'}} (u_{2}) = 3 (- 0.90027 \dots)^{2} - 2 = 0.43148 \dots

u_{3} = 10.52325 \dots

Δ u_{3} = ∣ 10.52325 \dots - (- 0.90027 \dots) ∣ = 11.42353 \dots

Δ u_{3} = 11.42353 \dots

u_{4} = 7.07616 \dots : Δ u_{4} = 3.44709 \dots

f (u_{3}) = 10.52325 \dots^{3} - 2 \cdot 10.52325 \dots - 6 = 1138.28856 \dots

f^{^{'}} (u_{3}) = 3 \cdot 10.52325 \dots^{2} - 2 = 330.21696 \dots

u_{4} = 7.07616 \dots

Δ u_{4} = ∣ 7.07616 \dots - 10.52325 \dots ∣ = 3.44709 \dots

Δ u_{4} = 3.44709 \dots

u_{5} = 4.82158 \dots : Δ u_{5} = 2.25458 \dots

f (u_{4}) = 7.07616 \dots^{3} - 2 \cdot 7.07616 \dots - 6 = 334.16639 \dots

f^{^{'}} (u_{4}) = 3 \cdot 7.07616 \dots^{2} - 2 = 148.21639 \dots

u_{5} = 4.82158 \dots

Δ u_{5} = ∣ 4.82158 \dots - 7.07616 \dots ∣ = 2.25458 \dots

Δ u_{5} = 2.25458 \dots

u_{6} = 3.39785 \dots : Δ u_{6} = 1.42372 \dots

f (u_{5}) = 4.82158 \dots^{3} - 2 \cdot 4.82158 \dots - 6 = 96.44728 \dots

f^{^{'}} (u_{5}) = 3 \cdot 4.82158 \dots^{2} - 2 = 67.74295 \dots

u_{6} = 3.39785 \dots

Δ u_{6} = ∣ 3.39785 \dots - 4.82158 \dots ∣ = 1.42372 \dots

Δ u_{6} = 1.42372 \dots

u_{7} = 2.58789 \dots : Δ u_{7} = 0.80995 \dots

f (u_{6}) = 3.39785 \dots^{3} - 2 \cdot 3.39785 \dots - 6 = 26.43402 \dots

f^{^{'}} (u_{6}) = 3 \cdot 3.39785 \dots^{2} - 2 = 32.63630 \dots

u_{7} = 2.58789 \dots

Δ u_{7} = ∣ 2.58789 \dots - 3.39785 \dots ∣ = 0.80995 \dots

Δ u_{7} = 0.80995 \dots

u_{8} = 2.24763 \dots : Δ u_{8} = 0.34026 \dots

f (u_{7}) = 2.58789 \dots^{3} - 2 \cdot 2.58789 \dots - 6 = 6.15594 \dots

f^{^{'}} (u_{7}) = 3 \cdot 2.58789 \dots^{2} - 2 = 18.09167 \dots

u_{8} = 2.24763 \dots

Δ u_{8} = ∣ 2.24763 \dots - 2.58789 \dots ∣ = 0.34026 \dots

Δ u_{8} = 0.34026 \dots

u_{9} = 2.18230 \dots : Δ u_{9} = 0.06533 \dots

f (u_{8}) = 2.24763 \dots^{3} - 2 \cdot 2.24763 \dots - 6 = 0.85948 \dots

f^{^{'}} (u_{8}) = 3 \cdot 2.24763 \dots^{2} - 2 = 13.15559 \dots

u_{9} = 2.18230 \dots

Δ u_{9} = ∣ 2.18230 \dots - 2.24763 \dots ∣ = 0.06533 \dots

Δ u_{9} = 0.06533 \dots

u_{10} = 2.17998 \dots : Δ u_{10} = 0.00231 \dots

f (u_{9}) = 2.18230 \dots^{3} - 2 \cdot 2.18230 \dots - 6 = 0.02850 \dots

f^{^{'}} (u_{9}) = 3 \cdot 2.18230 \dots^{2} - 2 = 12.28734 \dots

u_{10} = 2.17998 \dots

Δ u_{10} = ∣ 2.17998 \dots - 2.18230 \dots ∣ = 0.00231 \dots

Δ u_{10} = 0.00231 \dots

u_{11} = 2.17998 \dots : Δ u_{11} = 2.87294 E - 6

f (u_{10}) = 2.17998 \dots^{3} - 2 \cdot 2.17998 \dots - 6 = 0.00003 \dots

f^{^{'}} (u_{10}) = 3 \cdot 2.17998 \dots^{2} - 2 = 12.25699 \dots

u_{11} = 2.17998 \dots

Δ u_{11} = ∣ 2.17998 \dots - 2.17998 \dots ∣ = 2.87294 E - 6

Δ u_{11} = 2.87294 E - 6

u_{12} = 2.17998 \dots : Δ u_{12} = 4.4041 E - 12

f (u_{11}) = 2.17998 \dots^{3} - 2 \cdot 2.17998 \dots - 6 = 5.39808 E - 11

f^{^{'}} (u_{11}) = 3 \cdot 2.17998 \dots^{2} - 2 = 12.25695 \dots

u_{12} = 2.17998 \dots

Δ u_{12} = ∣ 2.17998 \dots - 2.17998 \dots ∣ = 4.4041 E - 12

Δ u_{12} = 4.4041 E - 12

u \approx 2.17998 \dots

使用长除法

Eq u a t i o n 0 : \frac{u ^{3} - 2 u - 6}{u - 2.17998 \dots} = u^{2} + 2.17998 \dots u + 2.75231 \dots

u^{2} + 2.17998 \dots u + 2.75231 \dots \approx 0

使用牛顿-拉弗森方法找到

u^{2} + 2.17998 \dots u + 2.75231 \dots = 0

的一个解:

u \in R

无解

u^{2} + 2.17998 \dots u + 2.75231 \dots = 0

牛顿-拉弗森近似法定义

f (u) = u^{2} + 2.17998 \dots u + 2.75231 \dots

找到

f^{^{'}} (u) : 2 u + 2.17998 \dots

\frac{d}{d u} (u^{2} + 2.17998 \dots u + 2.75231 \dots)

使用微分加减法定则:

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

= \frac{d}{d u} (u^{2}) + \frac{d}{d u} (2.17998 \dots u) + \frac{d}{d u} (2.75231 \dots)

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

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

使用幂法则:

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

= 2 u^{2 - 1}

化简

= 2 u

\frac{d}{d u} (2.17998 \dots u) = 2.17998 \dots

\frac{d}{d u} (2.17998 \dots u)

将常数提出:

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

= 2.17998 \dots \frac{d u}{d u}

使用常见微分定则:

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

= 2.17998 \dots \cdot 1

化简

= 2.17998 \dots

\frac{d}{d u} (2.75231 \dots) = 0

\frac{d}{d u} (2.75231 \dots)

常数微分:

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

= 0

= 2 u + 2.17998 \dots + 0

化简

= 2 u + 2.17998 \dots

令

u_{0} = - 1

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = - 9.73612 \dots : Δ u_{1} = 8.73612 \dots

f (u_{0}) = (- 1)^{2} + 2.17998 \dots (- 1) + 2.75231 \dots = 1.57233 \dots

f^{^{'}} (u_{0}) = 2 (- 1) + 2.17998 \dots = 0.17998 \dots

u_{1} = - 9.73612 \dots

Δ u_{1} = ∣ - 9.73612 \dots - (- 1) ∣ = 8.73612 \dots

Δ u_{1} = 8.73612 \dots

u_{2} = - 5.32259 \dots : Δ u_{2} = 4.41352 \dots

f (u_{1}) = (- 9.73612 \dots)^{2} + 2.17998 \dots (- 9.73612 \dots) + 2.75231 \dots = 76.31980 \dots

f^{^{'}} (u_{1}) = 2 (- 9.73612 \dots) + 2.17998 \dots = - 17.29226 \dots

u_{2} = - 5.32259 \dots

Δ u_{2} = ∣ - 5.32259 \dots - (- 9.73612 \dots) ∣ = 4.41352 \dots

Δ u_{2} = 4.41352 \dots

u_{3} = - 3.02150 \dots : Δ u_{3} = 2.30108 \dots

f (u_{2}) = (- 5.32259 \dots)^{2} + 2.17998 \dots (- 5.32259 \dots) + 2.75231 \dots = 19.47919 \dots

f^{^{'}} (u_{2}) = 2 (- 5.32259 \dots) + 2.17998 \dots = - 8.46521 \dots

u_{3} = - 3.02150 \dots

Δ u_{3} = ∣ - 3.02150 \dots - (- 5.32259 \dots) ∣ = 2.30108 \dots

Δ u_{3} = 2.30108 \dots

u_{4} = - 1.65082 \dots : Δ u_{4} = 1.37068 \dots

f (u_{3}) = (- 3.02150 \dots)^{2} + 2.17998 \dots (- 3.02150 \dots) + 2.75231 \dots = 5.29500 \dots

f^{^{'}} (u_{3}) = 2 (- 3.02150 \dots) + 2.17998 \dots = - 3.86303 \dots

u_{4} = - 1.65082 \dots

Δ u_{4} = ∣ - 1.65082 \dots - (- 3.02150 \dots) ∣ = 1.37068 \dots

Δ u_{4} = 1.37068 \dots

u_{5} = 0.02415 \dots : Δ u_{5} = 1.67497 \dots

f (u_{4}) = (- 1.65082 \dots)^{2} + 2.17998 \dots (- 1.65082 \dots) + 2.75231 \dots = 1.87877 \dots

f^{^{'}} (u_{4}) = 2 (- 1.65082 \dots) + 2.17998 \dots = - 1.12167 \dots

u_{5} = 0.02415 \dots

Δ u_{5} = ∣ 0.02415 \dots - (- 1.65082 \dots) ∣ = 1.67497 \dots

Δ u_{5} = 1.67497 \dots

u_{6} = - 1.23490 \dots : Δ u_{6} = 1.25906 \dots

f (u_{5}) = 0.02415 \dots^{2} + 2.17998 \dots \cdot 0.02415 \dots + 2.75231 \dots = 2.80555 \dots

f^{^{'}} (u_{5}) = 2 \cdot 0.02415 \dots + 2.17998 \dots = 2.22828 \dots

u_{6} = - 1.23490 \dots

Δ u_{6} = ∣ - 1.23490 \dots - 0.02415 \dots ∣ = 1.25906 \dots

Δ u_{6} = 1.25906 \dots

u_{7} = 4.23449 \dots : Δ u_{7} = 5.46939 \dots

f (u_{6}) = (- 1.23490 \dots)^{2} + 2.17998 \dots (- 1.23490 \dots) + 2.75231 \dots = 1.58523 \dots

f^{^{'}} (u_{6}) = 2 (- 1.23490 \dots) + 2.17998 \dots = - 0.28983 \dots

u_{7} = 4.23449 \dots

Δ u_{7} = ∣ 4.23449 \dots - (- 1.23490 \dots) ∣ = 5.46939 \dots

Δ u_{7} = 5.46939 \dots

u_{8} = 1.42535 \dots : Δ u_{8} = 2.80913 \dots

f (u_{7}) = 4.23449 \dots^{2} + 2.17998 \dots \cdot 4.23449 \dots + 2.75231 \dots = 29.91433 \dots

f^{^{'}} (u_{7}) = 2 \cdot 4.23449 \dots + 2.17998 \dots = 10.64896 \dots

u_{8} = 1.42535 \dots

Δ u_{8} = ∣ 1.42535 \dots - 4.23449 \dots ∣ = 2.80913 \dots

Δ u_{8} = 2.80913 \dots

u_{9} = - 0.14325 \dots : Δ u_{9} = 1.56861 \dots

f (u_{8}) = 1.42535 \dots^{2} + 2.17998 \dots \cdot 1.42535 \dots + 2.75231 \dots = 7.89122 \dots

f^{^{'}} (u_{8}) = 2 \cdot 1.42535 \dots + 2.17998 \dots = 5.03069 \dots

u_{9} = - 0.14325 \dots

Δ u_{9} = ∣ - 0.14325 \dots - 1.42535 \dots ∣ = 1.56861 \dots

Δ u_{9} = 1.56861 \dots

u_{10} = - 1.44274 \dots : Δ u_{10} = 1.29948 \dots

f (u_{9}) = (- 0.14325 \dots)^{2} + 2.17998 \dots (- 0.14325 \dots) + 2.75231 \dots = 2.46054 \dots

f^{^{'}} (u_{9}) = 2 (- 0.14325 \dots) + 2.17998 \dots = 1.89347 \dots

u_{10} = - 1.44274 \dots

Δ u_{10} = ∣ - 1.44274 \dots - (- 0.14325 \dots) ∣ = 1.29948 \dots

Δ u_{10} = 1.29948 \dots

u_{11} = 0.95081 \dots : Δ u_{11} = 2.39356 \dots

f (u_{10}) = (- 1.44274 \dots)^{2} + 2.17998 \dots (- 1.44274 \dots) + 2.75231 \dots = 1.68867 \dots

f^{^{'}} (u_{10}) = 2 (- 1.44274 \dots) + 2.17998 \dots = - 0.70550 \dots

u_{11} = 0.95081 \dots

Δ u_{11} = ∣ 0.95081 \dots - (- 1.44274 \dots) ∣ = 2.39356 \dots

Δ u_{11} = 2.39356 \dots

u_{12} = - 0.45282 \dots : Δ u_{12} = 1.40364 \dots

f (u_{11}) = 0.95081 \dots^{2} + 2.17998 \dots \cdot 0.95081 \dots + 2.75231 \dots = 5.72913 \dots

f^{^{'}} (u_{11}) = 2 \cdot 0.95081 \dots + 2.17998 \dots = 4.08161 \dots

u_{12} = - 0.45282 \dots

Δ u_{12} = ∣ - 0.45282 \dots - 0.95081 \dots ∣ = 1.40364 \dots

Δ u_{12} = 1.40364 \dots

u_{13} = - 1.99891 \dots : Δ u_{13} = 1.54608 \dots

f (u_{12}) = (- 0.45282 \dots)^{2} + 2.17998 \dots (- 0.45282 \dots) + 2.75231 \dots = 1.97021 \dots

f^{^{'}} (u_{12}) = 2 (- 0.45282 \dots) + 2.17998 \dots = 1.27432 \dots

u_{13} = - 1.99891 \dots

Δ u_{13} = ∣ - 1.99891 \dots - (- 0.45282 \dots) ∣ = 1.54608 \dots

Δ u_{13} = 1.54608 \dots

无法得出解

解是

u \approx 2.17998 \dots

u \approx 2.17998 \dots

验证解

找到无定义的点（奇点）:

u = 0

取

- 2 + u^{2} - \frac{6}{u}

的分母，令其等于零

u = 0

以下点无定义

u = 0

将不在定义域的点与解相综合:

u \approx 2.17998 \dots

u = cot (x)

代回

cot (x) \approx 2.17998 \dots

cot (x) \approx 2.17998 \dots

cot (x) = 2.17998 \dots : x = arccot (2.17998 \dots) + πn

cot (x) = 2.17998 \dots

使用反三角函数性质

cot (x) = 2.17998 \dots

cot (x) = 2.17998 \dots

的通解

cot (x) = a \Rightarrow x = arccot (a) + πn

x = arccot (2.17998 \dots) + πn

x = arccot (2.17998 \dots) + πn

合并所有解

x = arccot (2.17998 \dots) + πn

合并所有解

x = arccot (- 2.17998 \dots) + πn, x = arccot (2.17998 \dots) + πn

将解代入原方程进行验证

将它们代入

csc^{2} (x) - 3 = 6 tan (x)

检验解是否符合

去除与方程不符的解。

检验

arccot (- 2.17998 \dots) + πn

的解:

假

arccot (- 2.17998 \dots) + πn

代入

n = 1

arccot (- 2.17998 \dots) + π 1

对于

csc^{2} (x) - 3 = 6 tan (x)

代入

x = arccot (- 2.17998 \dots) + π 1

csc^{2} (arccot (- 2.17998 \dots) + π 1) - 3 = 6 tan (arccot (- 2.17998 \dots) + π 1)

整理后得

2.75231 \dots = - 2.75231 \dots

\Rightarrow 假

检验

arccot (2.17998 \dots) + πn

的解:

真

arccot (2.17998 \dots) + πn

代入

n = 1

arccot (2.17998 \dots) + π 1

对于

csc^{2} (x) - 3 = 6 tan (x)

代入

x = arccot (2.17998 \dots) + π 1

csc^{2} (arccot (2.17998 \dots) + π 1) - 3 = 6 tan (arccot (2.17998 \dots) + π 1)

整理后得

2.75231 \dots = 2.75231 \dots

\Rightarrow 真

x = arccot (2.17998 \dots) + πn

以小数形式表示解

x = 0.43008 \dots + πn

作图

流行的例子

solvefor C,arctan((1/(2*pi*R*C))/f)=0

so l v e f or C, arctan (\frac{\frac{1}{2 \cdot π \cdot R \cdot C}}{f}) = 0

sin(x)=-2cos(x)

sin (x) = - 2 cos (x)

3sin^2(x)-4cos(x)+1=0

3 sin^{2} (x) - 4 cos (x) + 1 = 0

sin(x)-cos(2x)-1=0

sin (x) - cos (2 x) - 1 = 0

sin^2(x)=sqrt(3)

sin^{2} (x) = 3