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受欢迎的三角函数 >

cos^3(x)-2sin(x)-0.7=0

解答

cos^{3} (x) - 2 sin (x) - 0.7 = 0

解答

x = 0.13658 \dots + 2 πn, x = - 2.49372 \dots + 2 πn

+1

度数

x = 7.82570 \dots^{\circ} + 36 0^{\circ} n, x = - 142.87998 \dots^{\circ} + 36 0^{\circ} n

求解步骤

cos^{3} (x) - 2 sin (x) - 0.7 = 0

两边加上

2 sin (x)

cos^{3} (x) - 0.7 = 2 sin (x)

两边进行平方

(cos^{3} (x) - 0.7)^{2} = (2 sin (x))^{2}

两边减去

(2 sin (x))^{2}

(cos^{3} (x) - 0.7)^{2} - 4 sin^{2} (x) = 0

使用三角恒等式改写

(- 0.7 + cos^{3} (x))^{2} - 4 sin^{2} (x)

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= (- 0.7 + cos^{3} (x))^{2} - 4 (1 - cos^{2} (x))

化简

(- 0.7 + cos^{3} (x))^{2} - 4 (1 - cos^{2} (x)) : cos^{6} (x) + 4 cos^{2} (x) - 1.4 cos^{3} (x) - 3.51

(- 0.7 + cos^{3} (x))^{2} - 4 (1 - cos^{2} (x))

(- 0.7 + cos^{3} (x))^{2} : 0.49 - 1.4 cos^{3} (x) + cos^{6} (x)

使用完全平方公式:

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

a = - 0.7, b = cos^{3} (x)

= (- 0.7)^{2} + 2 (- 0.7) cos^{3} (x) + (cos^{3} (x))^{2}

化简

(- 0.7)^{2} + 2 (- 0.7) cos^{3} (x) + (cos^{3} (x))^{2} : 0.49 - 1.4 cos^{3} (x) + cos^{6} (x)

(- 0.7)^{2} + 2 (- 0.7) cos^{3} (x) + (cos^{3} (x))^{2}

去除括号:

(- a) = - a

= (- 0.7)^{2} - 2 \cdot 0.7 cos^{3} (x) + (cos^{3} (x))^{2}

(- 0.7)^{2} = 0.49

(- 0.7)^{2}

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 0.7)^{2} = 0. 7^{2}

= 0. 7^{2}

0. 7^{2} = 0.49

= 0.49

2 \cdot 0.7 cos^{3} (x) = 1.4 cos^{3} (x)

2 \cdot 0.7 cos^{3} (x)

数字相乘:

2 \cdot 0.7 = 1.4

= 1.4 cos^{3} (x)

(cos^{3} (x))^{2} = cos^{6} (x)

(cos^{3} (x))^{2}

使用指数法则:

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

= cos^{3 \cdot 2} (x)

数字相乘:

3 \cdot 2 = 6

= cos^{6} (x)

= 0.49 - 1.4 cos^{3} (x) + cos^{6} (x)

= 0.49 - 1.4 cos^{3} (x) + cos^{6} (x)

= 0.49 - 1.4 cos^{3} (x) + cos^{6} (x) - 4 (1 - cos^{2} (x))

乘开

- 4 (1 - cos^{2} (x)) : - 4 + 4 cos^{2} (x)

- 4 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = - 4, b = 1, c = cos^{2} (x)

= - 4 \cdot 1 - (- 4) cos^{2} (x)

使用加减运算法则

- (- a) = a

= - 4 \cdot 1 + 4 cos^{2} (x)

数字相乘:

4 \cdot 1 = 4

= - 4 + 4 cos^{2} (x)

= 0.49 - 1.4 cos^{3} (x) + cos^{6} (x) - 4 + 4 cos^{2} (x)

化简

0.49 - 1.4 cos^{3} (x) + cos^{6} (x) - 4 + 4 cos^{2} (x) : cos^{6} (x) + 4 cos^{2} (x) - 1.4 cos^{3} (x) - 3.51

0.49 - 1.4 cos^{3} (x) + cos^{6} (x) - 4 + 4 cos^{2} (x)

对同类项分组

= - 1.4 cos^{3} (x) + cos^{6} (x) + 4 cos^{2} (x) + 0.49 - 4

数字相加/相减:

0.49 - 4 = - 3.51

= cos^{6} (x) + 4 cos^{2} (x) - 1.4 cos^{3} (x) - 3.51

= cos^{6} (x) + 4 cos^{2} (x) - 1.4 cos^{3} (x) - 3.51

= cos^{6} (x) + 4 cos^{2} (x) - 1.4 cos^{3} (x) - 3.51

- 3.51 + cos^{6} (x) - 1.4 cos^{3} (x) + 4 cos^{2} (x) = 0

用替代法求解

- 3.51 + cos^{6} (x) - 1.4 cos^{3} (x) + 4 cos^{2} (x) = 0

令

: cos (x) = u

- 3.51 + u^{6} - 1.4 u^{3} + 4 u^{2} = 0

- 3.51 + u^{6} - 1.4 u^{3} + 4 u^{2} = 0 : u \approx 0.99068 \dots, u \approx - 0.79737 \dots

- 3.51 + u^{6} - 1.4 u^{3} + 4 u^{2} = 0

在两边乘以

100

- 3.51 + u^{6} - 1.4 u^{3} + 4 u^{2} = 0

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere are

2

digits to the right of the decimal point, therefore multiply by

100

- 3.51 \cdot 100 + u^{6} \cdot 100 - 1.4 u^{3} \cdot 100 + 4 u^{2} \cdot 100 = 0 \cdot 100

整理后得

- 351 + 100 u^{6} - 140 u^{3} + 400 u^{2} = 0

- 351 + 100 u^{6} - 140 u^{3} + 400 u^{2} = 0

改写成标准形式

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

100 u^{6} - 140 u^{3} + 400 u^{2} - 351 = 0

使用牛顿-拉弗森方法找到

100 u^{6} - 140 u^{3} + 400 u^{2} - 351 = 0

的一个解:

u \approx 0.99068 \dots

100 u^{6} - 140 u^{3} + 400 u^{2} - 351 = 0

牛顿-拉弗森近似法定义

f (u) = 100 u^{6} - 140 u^{3} + 400 u^{2} - 351

找到

f^{^{'}} (u) : 600 u^{5} - 420 u^{2} + 800 u

\frac{d}{d u} (100 u^{6} - 140 u^{3} + 400 u^{2} - 351)

使用微分加减法定则:

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

= \frac{d}{d u} (100 u^{6}) - \frac{d}{d u} (140 u^{3}) + \frac{d}{d u} (400 u^{2}) - \frac{d}{d u} (351)

\frac{d}{d u} (100 u^{6}) = 600 u^{5}

\frac{d}{d u} (100 u^{6})

将常数提出:

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

= 100 \frac{d}{d u} (u^{6})

使用幂法则:

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

= 100 \cdot 6 u^{6 - 1}

化简

= 600 u^{5}

\frac{d}{d u} (140 u^{3}) = 420 u^{2}

\frac{d}{d u} (140 u^{3})

将常数提出:

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

= 140 \frac{d}{d u} (u^{3})

使用幂法则:

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

= 140 \cdot 3 u^{3 - 1}

化简

= 420 u^{2}

\frac{d}{d u} (400 u^{2}) = 800 u

\frac{d}{d u} (400 u^{2})

将常数提出:

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

= 400 \frac{d}{d u} (u^{2})

使用幂法则:

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

= 400 \cdot 2 u^{2 - 1}

化简

= 800 u

\frac{d}{d u} (351) = 0

\frac{d}{d u} (351)

常数微分:

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

= 0

= 600 u^{5} - 420 u^{2} + 800 u - 0

化简

= 600 u^{5} - 420 u^{2} + 800 u

令

u_{0} = 1

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 0.99081 \dots : Δ u_{1} = 0.00918 \dots

f (u_{0}) = 100 \cdot 1^{6} - 140 \cdot 1^{3} + 400 \cdot 1^{2} - 351 = 9

f^{^{'}} (u_{0}) = 600 \cdot 1^{5} - 420 \cdot 1^{2} + 800 \cdot 1 = 980

u_{1} = 0.99081 \dots

Δ u_{1} = ∣ 0.99081 \dots - 1 ∣ = 0.00918 \dots

Δ u_{1} = 0.00918 \dots

u_{2} = 0.99068 \dots : Δ u_{2} = 0.00012 \dots

f (u_{1}) = 100 \cdot 0.99081 \dots^{6} - 140 \cdot 0.99081 \dots^{3} + 400 \cdot 0.99081 \dots^{2} - 351 = 0.12339 \dots

f^{^{'}} (u_{1}) = 600 \cdot 0.99081 \dots^{5} - 420 \cdot 0.99081 \dots^{2} + 800 \cdot 0.99081 \dots = 953.28231 \dots

u_{2} = 0.99068 \dots

Δ u_{2} = ∣ 0.99068 \dots - 0.99081 \dots ∣ = 0.00012 \dots

Δ u_{2} = 0.00012 \dots

u_{3} = 0.99068 \dots : Δ u_{3} = 2.51305 E - 8

f (u_{2}) = 100 \cdot 0.99068 \dots^{6} - 140 \cdot 0.99068 \dots^{3} + 400 \cdot 0.99068 \dots^{2} - 351 = 0.00002 \dots

f^{^{'}} (u_{2}) = 600 \cdot 0.99068 \dots^{5} - 420 \cdot 0.99068 \dots^{2} + 800 \cdot 0.99068 \dots = 952.91233 \dots

u_{3} = 0.99068 \dots

Δ u_{3} = ∣ 0.99068 \dots - 0.99068 \dots ∣ = 2.51305 E - 8

Δ u_{3} = 2.51305 E - 8

u \approx 0.99068 \dots

使用长除法

Eq u a t i o n 0 : \frac{100 u ^{6} - 140 u ^{3} + 400 u ^{2} - 351}{u - 0.99068 \dots} = 100 u^{5} + 99.06868 \dots u^{4} + 98.14604 \dots u^{3} - 42.76800 \dots u^{2} + 357.63030 \dots u + 354.29964 \dots

100 u^{5} + 99.06868 \dots u^{4} + 98.14604 \dots u^{3} - 42.76800 \dots u^{2} + 357.63030 \dots u + 354.29964 \dots \approx 0

使用牛顿-拉弗森方法找到

100 u^{5} + 99.06868 \dots u^{4} + 98.14604 \dots u^{3} - 42.76800 \dots u^{2} + 357.63030 \dots u + 354.29964 \dots = 0

的一个解:

u \approx - 0.79737 \dots

100 u^{5} + 99.06868 \dots u^{4} + 98.14604 \dots u^{3} - 42.76800 \dots u^{2} + 357.63030 \dots u + 354.29964 \dots = 0

牛顿-拉弗森近似法定义

f (u) = 100 u^{5} + 99.06868 \dots u^{4} + 98.14604 \dots u^{3} - 42.76800 \dots u^{2} + 357.63030 \dots u + 354.29964 \dots

找到

f^{^{'}} (u) : 500 u^{4} + 396.27474 \dots u^{3} + 294.43813 \dots u^{2} - 85.53600 \dots u + 357.63030 \dots

\frac{d}{d u} (100 u^{5} + 99.06868 \dots u^{4} + 98.14604 \dots u^{3} - 42.76800 \dots u^{2} + 357.63030 \dots u + 354.29964 \dots)

使用微分加减法定则:

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

= \frac{d}{d u} (100 u^{5}) + \frac{d}{d u} (99.06868 \dots u^{4}) + \frac{d}{d u} (98.14604 \dots u^{3}) - \frac{d}{d u} (42.76800 \dots u^{2}) + \frac{d}{d u} (357.63030 \dots u) + \frac{d}{d u} (354.29964 \dots)

\frac{d}{d u} (100 u^{5}) = 500 u^{4}

\frac{d}{d u} (100 u^{5})

将常数提出:

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

= 100 \frac{d}{d u} (u^{5})

使用幂法则:

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

= 100 \cdot 5 u^{5 - 1}

化简

= 500 u^{4}

\frac{d}{d u} (99.06868 \dots u^{4}) = 396.27474 \dots u^{3}

\frac{d}{d u} (99.06868 \dots u^{4})

将常数提出:

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

= 99.06868 \dots \frac{d}{d u} (u^{4})

使用幂法则:

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

= 99.06868 \dots \cdot 4 u^{4 - 1}

化简

= 396.27474 \dots u^{3}

\frac{d}{d u} (98.14604 \dots u^{3}) = 294.43813 \dots u^{2}

\frac{d}{d u} (98.14604 \dots u^{3})

将常数提出:

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

= 98.14604 \dots \frac{d}{d u} (u^{3})

使用幂法则:

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

= 98.14604 \dots \cdot 3 u^{3 - 1}

化简

= 294.43813 \dots u^{2}

\frac{d}{d u} (42.76800 \dots u^{2}) = 85.53600 \dots u

\frac{d}{d u} (42.76800 \dots u^{2})

将常数提出:

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

= 42.76800 \dots \frac{d}{d u} (u^{2})

使用幂法则:

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

= 42.76800 \dots \cdot 2 u^{2 - 1}

化简

= 85.53600 \dots u

\frac{d}{d u} (357.63030 \dots u) = 357.63030 \dots

\frac{d}{d u} (357.63030 \dots u)

将常数提出:

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

= 357.63030 \dots \frac{d u}{d u}

使用常见微分定则:

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

= 357.63030 \dots \cdot 1

化简

= 357.63030 \dots

\frac{d}{d u} (354.29964 \dots) = 0

\frac{d}{d u} (354.29964 \dots)

常数微分:

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

= 0

= 500 u^{4} + 396.27474 \dots u^{3} + 294.43813 \dots u^{2} - 85.53600 \dots u + 357.63030 \dots + 0

化简

= 500 u^{4} + 396.27474 \dots u^{3} + 294.43813 \dots u^{2} - 85.53600 \dots u + 357.63030 \dots

令

u_{0} = - 1

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = - 0.82744 \dots : Δ u_{1} = 0.17255 \dots

f (u_{0}) = 100 (- 1)^{5} + 99.06868 \dots (- 1)^{4} + 98.14604 \dots (- 1)^{3} - 42.76800 \dots (- 1)^{2} + 357.63030 \dots (- 1) + 354.29964 \dots = - 145.17602 \dots

f^{^{'}} (u_{0}) = 500 (- 1)^{4} + 396.27474 \dots (- 1)^{3} + 294.43813 \dots (- 1)^{2} - 85.53600 \dots (- 1) + 357.63030 \dots = 841.32969 \dots

u_{1} = - 0.82744 \dots

Δ u_{1} = ∣ - 0.82744 \dots - (- 1) ∣ = 0.17255 \dots

Δ u_{1} = 0.17255 \dots

u_{2} = - 0.79798 \dots : Δ u_{2} = 0.02946 \dots

f (u_{1}) = 100 (- 0.82744 \dots)^{5} + 99.06868 \dots (- 0.82744 \dots)^{4} + 98.14604 \dots (- 0.82744 \dots)^{3} - 42.76800 \dots (- 0.82744 \dots)^{2} + 357.63030 \dots (- 0.82744 \dots) + 354.29964 \dots = - 18.85098 \dots

f^{^{'}} (u_{1}) = 500 (- 0.82744 \dots)^{4} + 396.27474 \dots (- 0.82744 \dots)^{3} + 294.43813 \dots (- 0.82744 \dots)^{2} - 85.53600 \dots (- 0.82744 \dots) + 357.63030 \dots = 639.88234 \dots

u_{2} = - 0.79798 \dots

Δ u_{2} = ∣ - 0.79798 \dots - (- 0.82744 \dots) ∣ = 0.02946 \dots

Δ u_{2} = 0.02946 \dots

u_{3} = - 0.79737 \dots : Δ u_{3} = 0.00061 \dots

f (u_{2}) = 100 (- 0.79798 \dots)^{5} + 99.06868 \dots (- 0.79798 \dots)^{4} + 98.14604 \dots (- 0.79798 \dots)^{3} - 42.76800 \dots (- 0.79798 \dots)^{2} + 357.63030 \dots (- 0.79798 \dots) + 354.29964 \dots = - 0.37564 \dots

f^{^{'}} (u_{2}) = 500 (- 0.79798 \dots)^{4} + 396.27474 \dots (- 0.79798 \dots)^{3} + 294.43813 \dots (- 0.79798 \dots)^{2} - 85.53600 \dots (- 0.79798 \dots) + 357.63030 \dots = 614.75965 \dots

u_{3} = - 0.79737 \dots

Δ u_{3} = ∣ - 0.79737 \dots - (- 0.79798 \dots) ∣ = 0.00061 \dots

Δ u_{3} = 0.00061 \dots

u_{4} = - 0.79737 \dots : Δ u_{4} = 2.47446 E - 7

f (u_{3}) = 100 (- 0.79737 \dots)^{5} + 99.06868 \dots (- 0.79737 \dots)^{4} + 98.14604 \dots (- 0.79737 \dots)^{3} - 42.76800 \dots (- 0.79737 \dots)^{2} + 357.63030 \dots (- 0.79737 \dots) + 354.29964 \dots = - 0.00015 \dots

f^{^{'}} (u_{3}) = 500 (- 0.79737 \dots)^{4} + 396.27474 \dots (- 0.79737 \dots)^{3} + 294.43813 \dots (- 0.79737 \dots)^{2} - 85.53600 \dots (- 0.79737 \dots) + 357.63030 \dots = 614.26230 \dots

u_{4} = - 0.79737 \dots

Δ u_{4} = ∣ - 0.79737 \dots - (- 0.79737 \dots) ∣ = 2.47446 E - 7

Δ u_{4} = 2.47446 E - 7

u \approx - 0.79737 \dots

使用长除法

Eq u a t i o n 0 : \frac{100 u ^{5} + 99.06868 \dots u ^{4} + 98.14604 \dots u ^{3} - 42.76800 \dots u ^{2} + 357.63030 \dots u + 354.29964 \dots}{u + 0.79737 \dots} = 100 u^{4} + 19.33136 \dots u^{3} + 82.73173 \dots u^{2} - 108.73606 \dots u + 444.33352 \dots

100 u^{4} + 19.33136 \dots u^{3} + 82.73173 \dots u^{2} - 108.73606 \dots u + 444.33352 \dots \approx 0

使用牛顿-拉弗森方法找到

100 u^{4} + 19.33136 \dots u^{3} + 82.73173 \dots u^{2} - 108.73606 \dots u + 444.33352 \dots = 0

的一个解:

u \in R

无解

100 u^{4} + 19.33136 \dots u^{3} + 82.73173 \dots u^{2} - 108.73606 \dots u + 444.33352 \dots = 0

牛顿-拉弗森近似法定义

f (u) = 100 u^{4} + 19.33136 \dots u^{3} + 82.73173 \dots u^{2} - 108.73606 \dots u + 444.33352 \dots

找到

f^{^{'}} (u) : 400 u^{3} + 57.99410 \dots u^{2} + 165.46346 \dots u - 108.73606 \dots

\frac{d}{d u} (100 u^{4} + 19.33136 \dots u^{3} + 82.73173 \dots u^{2} - 108.73606 \dots u + 444.33352 \dots)

使用微分加减法定则:

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

= \frac{d}{d u} (100 u^{4}) + \frac{d}{d u} (19.33136 \dots u^{3}) + \frac{d}{d u} (82.73173 \dots u^{2}) - \frac{d}{d u} (108.73606 \dots u) + \frac{d}{d u} (444.33352 \dots)

\frac{d}{d u} (100 u^{4}) = 400 u^{3}

\frac{d}{d u} (100 u^{4})

将常数提出:

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

= 100 \frac{d}{d u} (u^{4})

使用幂法则:

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

= 100 \cdot 4 u^{4 - 1}

化简

= 400 u^{3}

\frac{d}{d u} (19.33136 \dots u^{3}) = 57.99410 \dots u^{2}

\frac{d}{d u} (19.33136 \dots u^{3})

将常数提出:

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

= 19.33136 \dots \frac{d}{d u} (u^{3})

使用幂法则:

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

= 19.33136 \dots \cdot 3 u^{3 - 1}

化简

= 57.99410 \dots u^{2}

\frac{d}{d u} (82.73173 \dots u^{2}) = 165.46346 \dots u

\frac{d}{d u} (82.73173 \dots u^{2})

将常数提出:

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

= 82.73173 \dots \frac{d}{d u} (u^{2})

使用幂法则:

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

= 82.73173 \dots \cdot 2 u^{2 - 1}

化简

= 165.46346 \dots u

\frac{d}{d u} (108.73606 \dots u) = 108.73606 \dots

\frac{d}{d u} (108.73606 \dots u)

将常数提出:

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

= 108.73606 \dots \frac{d u}{d u}

使用常见微分定则:

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

= 108.73606 \dots \cdot 1

化简

= 108.73606 \dots

\frac{d}{d u} (444.33352 \dots) = 0

\frac{d}{d u} (444.33352 \dots)

常数微分:

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

= 0

= 400 u^{3} + 57.99410 \dots u^{2} + 165.46346 \dots u - 108.73606 \dots + 0

化简

= 400 u^{3} + 57.99410 \dots u^{2} + 165.46346 \dots u - 108.73606 \dots

令

u_{0} = 4

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 2.95977 \dots : Δ u_{1} = 1.04022 \dots

f (u_{0}) = 100 \cdot 4^{4} + 19.33136 \dots \cdot 4^{3} + 82.73173 \dots \cdot 4^{2} - 108.73606 \dots \cdot 4 + 444.33352 \dots = 28170.30446 \dots

f^{^{'}} (u_{0}) = 400 \cdot 4^{3} + 57.99410 \dots \cdot 4^{2} + 165.46346 \dots \cdot 4 - 108.73606 \dots = 27081.02341 \dots

u_{1} = 2.95977 \dots

Δ u_{1} = ∣ 2.95977 \dots - 4 ∣ = 1.04022 \dots

Δ u_{1} = 1.04022 \dots

u_{2} = 2.15849 \dots : Δ u_{2} = 0.80127 \dots

f (u_{1}) = 100 \cdot 2.95977 \dots^{4} + 19.33136 \dots \cdot 2.95977 \dots^{3} + 82.73173 \dots \cdot 2.95977 \dots^{2} - 108.73606 \dots \cdot 2.95977 \dots + 444.33352 \dots = 9022.73494 \dots

f^{^{'}} (u_{1}) = 400 \cdot 2.95977 \dots^{3} + 57.99410 \dots \cdot 2.95977 \dots^{2} + 165.46346 \dots \cdot 2.95977 \dots - 108.73606 \dots = 11260.43299 \dots

u_{2} = 2.15849 \dots

Δ u_{2} = ∣ 2.15849 \dots - 2.95977 \dots ∣ = 0.80127 \dots

Δ u_{2} = 0.80127 \dots

u_{3} = 1.50665 \dots : Δ u_{3} = 0.65184 \dots

f (u_{2}) = 100 \cdot 2.15849 \dots^{4} + 19.33136 \dots \cdot 2.15849 \dots^{3} + 82.73173 \dots \cdot 2.15849 \dots^{2} - 108.73606 \dots \cdot 2.15849 \dots + 444.33352 \dots = 2960.23225 \dots

f^{^{'}} (u_{2}) = 400 \cdot 2.15849 \dots^{3} + 57.99410 \dots \cdot 2.15849 \dots^{2} + 165.46346 \dots \cdot 2.15849 \dots - 108.73606 \dots = 4541.29955 \dots

u_{3} = 1.50665 \dots

Δ u_{3} = ∣ 1.50665 \dots - 2.15849 \dots ∣ = 0.65184 \dots

Δ u_{3} = 0.65184 \dots

u_{4} = 0.86667 \dots : Δ u_{4} = 0.63997 \dots

f (u_{3}) = 100 \cdot 1.50665 \dots^{4} + 19.33136 \dots \cdot 1.50665 \dots^{3} + 82.73173 \dots \cdot 1.50665 \dots^{2} - 108.73606 \dots \cdot 1.50665 \dots + 444.33352 \dots = 1049.71285 \dots

f^{^{'}} (u_{3}) = 400 \cdot 1.50665 \dots^{3} + 57.99410 \dots \cdot 1.50665 \dots^{2} + 165.46346 \dots \cdot 1.50665 \dots - 108.73606 \dots = 1640.24724 \dots

u_{4} = 0.86667 \dots

Δ u_{4} = ∣ 0.86667 \dots - 1.50665 \dots ∣ = 0.63997 \dots

Δ u_{4} = 0.63997 \dots

u_{5} = - 0.55447 \dots : Δ u_{5} = 1.42115 \dots

f (u_{4}) = 100 \cdot 0.86667 \dots^{4} + 19.33136 \dots \cdot 0.86667 \dots^{3} + 82.73173 \dots \cdot 0.86667 \dots^{2} - 108.73606 \dots \cdot 0.86667 \dots + 444.33352 \dots = 481.24159 \dots

f^{^{'}} (u_{4}) = 400 \cdot 0.86667 \dots^{3} + 57.99410 \dots \cdot 0.86667 \dots^{2} + 165.46346 \dots \cdot 0.86667 \dots - 108.73606 \dots = 338.62623 \dots

u_{5} = - 0.55447 \dots

Δ u_{5} = ∣ - 0.55447 \dots - 0.86667 \dots ∣ = 1.42115 \dots

Δ u_{5} = 1.42115 \dots

u_{6} = 1.58320 \dots : Δ u_{6} = 2.13768 \dots

f (u_{5}) = 100 (- 0.55447 \dots)^{4} + 19.33136 \dots (- 0.55447 \dots)^{3} + 82.73173 \dots (- 0.55447 \dots)^{2} - 108.73606 \dots (- 0.55447 \dots) + 444.33352 \dots = 536.21783 \dots

f^{^{'}} (u_{5}) = 400 (- 0.55447 \dots)^{3} + 57.99410 \dots (- 0.55447 \dots)^{2} + 165.46346 \dots (- 0.55447 \dots) - 108.73606 \dots = - 250.84100 \dots

u_{6} = 1.58320 \dots

Δ u_{6} = ∣ 1.58320 \dots - (- 0.55447 \dots) ∣ = 2.13768 \dots

Δ u_{6} = 2.13768 \dots

u_{7} = 0.95511 \dots : Δ u_{7} = 0.62809 \dots

f (u_{6}) = 100 \cdot 1.58320 \dots^{4} + 19.33136 \dots \cdot 1.58320 \dots^{3} + 82.73173 \dots \cdot 1.58320 \dots^{2} - 108.73606 \dots \cdot 1.58320 \dots + 444.33352 \dots = 1184.53262 \dots

f^{^{'}} (u_{6}) = 400 \cdot 1.58320 \dots^{3} + 57.99410 \dots \cdot 1.58320 \dots^{2} + 165.46346 \dots \cdot 1.58320 \dots - 108.73606 \dots = 1885.92418 \dots

u_{7} = 0.95511 \dots

Δ u_{7} = ∣ 0.95511 \dots - 1.58320 \dots ∣ = 0.62809 \dots

Δ u_{7} = 0.62809 \dots

u_{8} = - 0.18975 \dots : Δ u_{8} = 1.14486 \dots

f (u_{7}) = 100 \cdot 0.95511 \dots^{4} + 19.33136 \dots \cdot 0.95511 \dots^{3} + 82.73173 \dots \cdot 0.95511 \dots^{2} - 108.73606 \dots \cdot 0.95511 \dots + 444.33352 \dots = 516.00984 \dots

f^{^{'}} (u_{7}) = 400 \cdot 0.95511 \dots^{3} + 57.99410 \dots \cdot 0.95511 \dots^{2} + 165.46346 \dots \cdot 0.95511 \dots - 108.73606 \dots = 450.71799 \dots

u_{8} = - 0.18975 \dots

Δ u_{8} = ∣ - 0.18975 \dots - 0.95511 \dots ∣ = 1.14486 \dots

Δ u_{8} = 1.14486 \dots

u_{9} = 3.13422 \dots : Δ u_{9} = 3.32398 \dots

f (u_{8}) = 100 (- 0.18975 \dots)^{4} + 19.33136 \dots (- 0.18975 \dots)^{3} + 82.73173 \dots (- 0.18975 \dots)^{2} - 108.73606 \dots (- 0.18975 \dots) + 444.33352 \dots = 467.94277 \dots

f^{^{'}} (u_{8}) = 400 (- 0.18975 \dots)^{3} + 57.99410 \dots (- 0.18975 \dots)^{2} + 165.46346 \dots (- 0.18975 \dots) - 108.73606 \dots = - 140.77780 \dots

u_{9} = 3.13422 \dots

Δ u_{9} = ∣ 3.13422 \dots - (- 0.18975 \dots) ∣ = 3.32398 \dots

Δ u_{9} = 3.32398 \dots

无法得出解

解为

u \approx 0.99068 \dots, u \approx - 0.79737 \dots

u = cos (x)

代回

cos (x) \approx 0.99068 \dots, cos (x) \approx - 0.79737 \dots

cos (x) \approx 0.99068 \dots, cos (x) \approx - 0.79737 \dots

cos (x) = 0.99068 \dots : x = arccos (0.99068 \dots) + 2 πn, x = 2 π - arccos (0.99068 \dots) + 2 πn

cos (x) = 0.99068 \dots

使用反三角函数性质

cos (x) = 0.99068 \dots

cos (x) = 0.99068 \dots

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (0.99068 \dots) + 2 πn, x = 2 π - arccos (0.99068 \dots) + 2 πn

x = arccos (0.99068 \dots) + 2 πn, x = 2 π - arccos (0.99068 \dots) + 2 πn

cos (x) = - 0.79737 \dots : x = arccos (- 0.79737 \dots) + 2 πn, x = - arccos (- 0.79737 \dots) + 2 πn

cos (x) = - 0.79737 \dots

使用反三角函数性质

cos (x) = - 0.79737 \dots

cos (x) = - 0.79737 \dots

的通解

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (- 0.79737 \dots) + 2 πn, x = - arccos (- 0.79737 \dots) + 2 πn

x = arccos (- 0.79737 \dots) + 2 πn, x = - arccos (- 0.79737 \dots) + 2 πn

合并所有解

x = arccos (0.99068 \dots) + 2 πn, x = 2 π - arccos (0.99068 \dots) + 2 πn, x = arccos (- 0.79737 \dots) + 2 πn, x = - arccos (- 0.79737 \dots) + 2 πn

将解代入原方程进行验证

将它们代入

cos^{3} (x) - 2 sin (x) - 0.7 = 0

检验解是否符合

去除与方程不符的解。

检验

arccos (0.99068 \dots) + 2 πn

的解:

真

arccos (0.99068 \dots) + 2 πn

代入

n = 1

arccos (0.99068 \dots) + 2 π 1

对于

cos^{3} (x) - 2 sin (x) - 0.7 = 0

代入

x = arccos (0.99068 \dots) + 2 π 1

cos^{3} (arccos (0.99068 \dots) + 2 π 1) - 2 sin (arccos (0.99068 \dots) + 2 π 1) - 0.7 = 0

整理后得

0 = 0

\Rightarrow 真

检验

2 π - arccos (0.99068 \dots) + 2 πn

的解:

假

2 π - arccos (0.99068 \dots) + 2 πn

代入

n = 1

2 π - arccos (0.99068 \dots) + 2 π 1

对于

cos^{3} (x) - 2 sin (x) - 0.7 = 0

代入

x = 2 π - arccos (0.99068 \dots) + 2 π 1

cos^{3} (2 π - arccos (0.99068 \dots) + 2 π 1) - 2 sin (2 π - arccos (0.99068 \dots) + 2 π 1) - 0.7 = 0

整理后得

0.54463 \dots = 0

\Rightarrow 假

检验

arccos (- 0.79737 \dots) + 2 πn

的解:

假

arccos (- 0.79737 \dots) + 2 πn

代入

n = 1

arccos (- 0.79737 \dots) + 2 π 1

对于

cos^{3} (x) - 2 sin (x) - 0.7 = 0

代入

x = arccos (- 0.79737 \dots) + 2 π 1

cos^{3} (arccos (- 0.79737 \dots) + 2 π 1) - 2 sin (arccos (- 0.79737 \dots) + 2 π 1) - 0.7 = 0

整理后得

- 2.41394 \dots = 0

\Rightarrow 假

检验

- arccos (- 0.79737 \dots) + 2 πn

的解:

真

- arccos (- 0.79737 \dots) + 2 πn

代入

n = 1

- arccos (- 0.79737 \dots) + 2 π 1

对于

cos^{3} (x) - 2 sin (x) - 0.7 = 0

代入

x = - arccos (- 0.79737 \dots) + 2 π 1

cos^{3} (- arccos (- 0.79737 \dots) + 2 π 1) - 2 sin (- arccos (- 0.79737 \dots) + 2 π 1) - 0.7 = 0

整理后得

0 = 0

\Rightarrow 真

x = arccos (0.99068 \dots) + 2 πn, x = - arccos (- 0.79737 \dots) + 2 πn

以小数形式表示解

x = 0.13658 \dots + 2 πn, x = - 2.49372 \dots + 2 πn

作图

流行的例子

solvefor y=cos(x),x

so l v e f or y = cos (x), x

tan(θ)= 5/(5sqrt(3))

tan (θ) = \frac{5}{5 3}

cos (x) = 7

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5 sin (θ) cos (θ) + 4 cos (θ) = 0

2 csc (x) + 4 = 0