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受欢迎的三角函数 >

2cot^2(x)+tan(x)=cot(x)

解答

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

解答

x = 2.15228 \dots + πn

+1

度数

x = 123.31684 \dots^{\circ} + 18 0^{\circ} n

求解步骤

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

两边减去

cot (x)

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

使用三角恒等式改写

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

使用基本三角恒等式:

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

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

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

用替代法求解

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

令

: cot (x) = u

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

- u + \frac{1}{u} + 2 u^{2} = 0 : u \approx - 0.65729 \dots

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

在两边乘以

u

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

在两边乘以

u

- uu + \frac{1}{u} u + 2 u^{2} u = 0 \cdot u

化简

- uu + \frac{1}{u} u + 2 u^{2} u = 0 \cdot u

化简

- uu : - u^{2}

- uu

使用指数法则:

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

uu = u^{1 + 1}

= - u^{1 + 1}

数字相加:

1 + 1 = 2

= - u^{2}

化简

\frac{1}{u} u : 1

\frac{1}{u} u

分式相乘:

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

= \frac{1 \cdot u}{u}

约分:

u

= 1

化简

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

2 u^{2} u

使用指数法则:

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

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

= 2 u^{2 + 1}

数字相加:

2 + 1 = 3

= 2 u^{3}

化简

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

- u^{2} + 1 + 2 u^{3} = 0

- u^{2} + 1 + 2 u^{3} = 0

- u^{2} + 1 + 2 u^{3} = 0

解

- u^{2} + 1 + 2 u^{3} = 0 : u \approx - 0.65729 \dots

- u^{2} + 1 + 2 u^{3} = 0

改写成标准形式

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

2 u^{3} - u^{2} + 1 = 0

使用牛顿-拉弗森方法找到

2 u^{3} - u^{2} + 1 = 0

的一个解:

u \approx - 0.65729 \dots

2 u^{3} - u^{2} + 1 = 0

牛顿-拉弗森近似法定义

f (u) = 2 u^{3} - u^{2} + 1

找到

f^{^{'}} (u) : 6 u^{2} - 2 u

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

使用微分加减法定则:

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

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

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

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

将常数提出:

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

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

使用幂法则:

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

= 2 \cdot 3 u^{3 - 1}

化简

= 6 u^{2}

\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} (1) = 0

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

常数微分:

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

= 0

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

化简

= 6 u^{2} - 2 u

令

u_{0} = - 1

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = - 0.75 : Δ u_{1} = 0.25

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

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

u_{1} = - 0.75

Δ u_{1} = ∣ - 0.75 - (- 1) ∣ = 0.25

Δ u_{1} = 0.25

u_{2} = - 0.66666 \dots : Δ u_{2} = 0.08333 \dots

f (u_{1}) = 2 (- 0.75)^{3} - (- 0.75)^{2} + 1 = - 0.40625

f^{^{'}} (u_{1}) = 6 (- 0.75)^{2} - 2 (- 0.75) = 4.875

u_{2} = - 0.66666 \dots

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

Δ u_{2} = 0.08333 \dots

u_{3} = - 0.65740 \dots : Δ u_{3} = 0.00925 \dots

f (u_{2}) = 2 (- 0.66666 \dots)^{3} - (- 0.66666 \dots)^{2} + 1 = - 0.03703 \dots

f^{^{'}} (u_{2}) = 6 (- 0.66666 \dots)^{2} - 2 (- 0.66666 \dots) = 4

u_{3} = - 0.65740 \dots

Δ u_{3} = ∣ - 0.65740 \dots - (- 0.66666 \dots) ∣ = 0.00925 \dots

Δ u_{3} = 0.00925 \dots

u_{4} = - 0.65729 \dots : Δ u_{4} = 0.00010 \dots

f (u_{3}) = 2 (- 0.65740 \dots)^{3} - (- 0.65740 \dots)^{2} + 1 = - 0.00042 \dots

f^{^{'}} (u_{3}) = 6 (- 0.65740 \dots)^{2} - 2 (- 0.65740 \dots) = 3.90792 \dots

u_{4} = - 0.65729 \dots

Δ u_{4} = ∣ - 0.65729 \dots - (- 0.65740 \dots) ∣ = 0.00010 \dots

Δ u_{4} = 0.00010 \dots

u_{5} = - 0.65729 \dots : Δ u_{5} = 1.51148 E - 8

f (u_{4}) = 2 (- 0.65729 \dots)^{3} - (- 0.65729 \dots)^{2} + 1 = - 5.90512 E - 8

f^{^{'}} (u_{4}) = 6 (- 0.65729 \dots)^{2} - 2 (- 0.65729 \dots) = 3.90684 \dots

u_{5} = - 0.65729 \dots

Δ u_{5} = ∣ - 0.65729 \dots - (- 0.65729 \dots) ∣ = 1.51148 E - 8

Δ u_{5} = 1.51148 E - 8

u \approx - 0.65729 \dots

使用长除法

Eq u a t i o n 0 : \frac{2 u ^{3} - u ^{2} + 1}{u + 0.65729 \dots} = 2 u^{2} - 2.31459 \dots u + 1.52137 \dots

2 u^{2} - 2.31459 \dots u + 1.52137 \dots \approx 0

使用牛顿-拉弗森方法找到

2 u^{2} - 2.31459 \dots u + 1.52137 \dots = 0

的一个解:

u \in R

无解

2 u^{2} - 2.31459 \dots u + 1.52137 \dots = 0

牛顿-拉弗森近似法定义

f (u) = 2 u^{2} - 2.31459 \dots u + 1.52137 \dots

找到

f^{^{'}} (u) : 4 u - 2.31459 \dots

\frac{d}{d u} (2 u^{2} - 2.31459 \dots u + 1.52137 \dots)

使用微分加减法定则:

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

= \frac{d}{d u} (2 u^{2}) - \frac{d}{d u} (2.31459 \dots u) + \frac{d}{d u} (1.52137 \dots)

\frac{d}{d u} (2 u^{2}) = 4 u

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

将常数提出:

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

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

使用幂法则:

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

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

化简

= 4 u

\frac{d}{d u} (2.31459 \dots u) = 2.31459 \dots

\frac{d}{d u} (2.31459 \dots u)

将常数提出:

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

= 2.31459 \dots \frac{d u}{d u}

使用常见微分定则:

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

= 2.31459 \dots \cdot 1

化简

= 2.31459 \dots

\frac{d}{d u} (1.52137 \dots) = 0

\frac{d}{d u} (1.52137 \dots)

常数微分:

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

= 0

= 4 u - 2.31459 \dots + 0

化简

= 4 u - 2.31459 \dots

令

u_{0} = 1

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 0.28397 \dots : Δ u_{1} = 0.71602 \dots

f (u_{0}) = 2 \cdot 1^{2} - 2.31459 \dots \cdot 1 + 1.52137 \dots = 1.20678 \dots

f^{^{'}} (u_{0}) = 4 \cdot 1 - 2.31459 \dots = 1.68540 \dots

u_{1} = 0.28397 \dots

Δ u_{1} = ∣ 0.28397 \dots - 1 ∣ = 0.71602 \dots

Δ u_{1} = 0.71602 \dots

u_{2} = 1.15391 \dots : Δ u_{2} = 0.86993 \dots

f (u_{1}) = 2 \cdot 0.28397 \dots^{2} - 2.31459 \dots \cdot 0.28397 \dots + 1.52137 \dots = 1.02537 \dots

f^{^{'}} (u_{1}) = 4 \cdot 0.28397 \dots - 2.31459 \dots = - 1.17867 \dots

u_{2} = 1.15391 \dots

Δ u_{2} = ∣ 1.15391 \dots - 0.28397 \dots ∣ = 0.86993 \dots

Δ u_{2} = 0.86993 \dots

u_{3} = 0.49614 \dots : Δ u_{3} = 0.65777 \dots

f (u_{2}) = 2 \cdot 1.15391 \dots^{2} - 2.31459 \dots \cdot 1.15391 \dots + 1.52137 \dots = 1.51356 \dots

f^{^{'}} (u_{2}) = 4 \cdot 1.15391 \dots - 2.31459 \dots = 2.30105 \dots

u_{3} = 0.49614 \dots

Δ u_{3} = ∣ 0.49614 \dots - 1.15391 \dots ∣ = 0.65777 \dots

Δ u_{3} = 0.65777 \dots

u_{4} = 3.11809 \dots : Δ u_{4} = 2.62195 \dots

f (u_{3}) = 2 \cdot 0.49614 \dots^{2} - 2.31459 \dots \cdot 0.49614 \dots + 1.52137 \dots = 0.86532 \dots

f^{^{'}} (u_{3}) = 4 \cdot 0.49614 \dots - 2.31459 \dots = - 0.33003 \dots

u_{4} = 3.11809 \dots

Δ u_{4} = ∣ 3.11809 \dots - 0.49614 \dots ∣ = 2.62195 \dots

Δ u_{4} = 2.62195 \dots

u_{5} = 1.76452 \dots : Δ u_{5} = 1.35357 \dots

f (u_{4}) = 2 \cdot 3.11809 \dots^{2} - 2.31459 \dots \cdot 3.11809 \dots + 1.52137 \dots = 13.74928 \dots

f^{^{'}} (u_{4}) = 4 \cdot 3.11809 \dots - 2.31459 \dots = 10.15778 \dots

u_{5} = 1.76452 \dots

Δ u_{5} = ∣ 1.76452 \dots - 3.11809 \dots ∣ = 1.35357 \dots

Δ u_{5} = 1.35357 \dots

u_{6} = 0.99203 \dots : Δ u_{6} = 0.77249 \dots

f (u_{5}) = 2 \cdot 1.76452 \dots^{2} - 2.31459 \dots \cdot 1.76452 \dots + 1.52137 \dots = 3.66430 \dots

f^{^{'}} (u_{5}) = 4 \cdot 1.76452 \dots - 2.31459 \dots = 4.74350 \dots

u_{6} = 0.99203 \dots

Δ u_{6} = ∣ 0.99203 \dots - 1.76452 \dots ∣ = 0.77249 \dots

Δ u_{6} = 0.77249 \dots

u_{7} = 0.27025 \dots : Δ u_{7} = 0.72177 \dots

f (u_{6}) = 2 \cdot 0.99203 \dots^{2} - 2.31459 \dots \cdot 0.99203 \dots + 1.52137 \dots = 1.19348 \dots

f^{^{'}} (u_{6}) = 4 \cdot 0.99203 \dots - 2.31459 \dots = 1.65353 \dots

u_{7} = 0.27025 \dots

Δ u_{7} = ∣ 0.27025 \dots - 0.99203 \dots ∣ = 0.72177 \dots

Δ u_{7} = 0.72177 \dots

u_{8} = 1.11489 \dots : Δ u_{8} = 0.84464 \dots

f (u_{7}) = 2 \cdot 0.27025 \dots^{2} - 2.31459 \dots \cdot 0.27025 \dots + 1.52137 \dots = 1.04192 \dots

f^{^{'}} (u_{7}) = 4 \cdot 0.27025 \dots - 2.31459 \dots = - 1.23356 \dots

u_{8} = 1.11489 \dots

Δ u_{8} = ∣ 1.11489 \dots - 0.27025 \dots ∣ = 0.84464 \dots

Δ u_{8} = 0.84464 \dots

u_{9} = 0.44970 \dots : Δ u_{9} = 0.66519 \dots

f (u_{8}) = 2 \cdot 1.11489 \dots^{2} - 2.31459 \dots \cdot 1.11489 \dots + 1.52137 \dots = 1.42683 \dots

f^{^{'}} (u_{8}) = 4 \cdot 1.11489 \dots - 2.31459 \dots = 2.14500 \dots

u_{9} = 0.44970 \dots

Δ u_{9} = ∣ 0.44970 \dots - 1.11489 \dots ∣ = 0.66519 \dots

Δ u_{9} = 0.66519 \dots

u_{10} = 2.16551 \dots : Δ u_{10} = 1.71580 \dots

f (u_{9}) = 2 \cdot 0.44970 \dots^{2} - 2.31459 \dots \cdot 0.44970 \dots + 1.52137 \dots = 0.88496 \dots

f^{^{'}} (u_{9}) = 4 \cdot 0.44970 \dots - 2.31459 \dots = - 0.51576 \dots

u_{10} = 2.16551 \dots

Δ u_{10} = ∣ 2.16551 \dots - 0.44970 \dots ∣ = 1.71580 \dots

Δ u_{10} = 1.71580 \dots

无法得出解

解是

u \approx - 0.65729 \dots

u \approx - 0.65729 \dots

验证解

找到无定义的点（奇点）:

u = 0

取

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

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 - 0.65729 \dots

u = cot (x)

代回

cot (x) \approx - 0.65729 \dots

cot (x) \approx - 0.65729 \dots

cot (x) = - 0.65729 \dots : x = arccot (- 0.65729 \dots) + πn

cot (x) = - 0.65729 \dots

使用反三角函数性质

cot (x) = - 0.65729 \dots

cot (x) = - 0.65729 \dots

的通解

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

x = arccot (- 0.65729 \dots) + πn

x = arccot (- 0.65729 \dots) + πn

合并所有解

x = arccot (- 0.65729 \dots) + πn

以小数形式表示解

x = 2.15228 \dots + πn

作图

流行的例子

sin(2x)=sin(pi/2-x)

sin (2 x) = sin (\frac{π}{2} - x)

tan (X) = 1

2sec^2(x)-sec(x)-1=0

2 sec^{2} (x) - sec (x) - 1 = 0

tan(2x-n/4)=-1,0<x<= 360

tan (2 x - \frac{n}{4}) = - 1, 0^{\circ} < x \leq 36 0^{\circ}

tan(θ)=-1,0<= θ<= 2pi

tan (θ) = - 1, 0 \leq θ \leq 2 π