zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

cos(3x)cos(x)=2cos(2x)cos(x)

解答

cos (3 x) cos (x) = 2 cos (2 x) cos (x)

解答

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = 1.04719 \dots + 2 πn, x = 2 π - 1.04719 \dots + 2 πn, x = 2.46670 \dots + 2 πn, x = - 2.46670 \dots + 2 πn

+1

度数

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n, x = 60.00000 \dots^{\circ} + 36 0^{\circ} n, x = 299.99999 \dots^{\circ} + 36 0^{\circ} n, x = 141.33171 \dots^{\circ} + 36 0^{\circ} n, x = - 141.33171 \dots^{\circ} + 36 0^{\circ} n

求解步骤

cos (3 x) cos (x) = 2 cos (2 x) cos (x)

两边减去

2 cos (2 x) cos (x)

cos (3 x) cos (x) - 2 cos (2 x) cos (x) = 0

使用三角恒等式改写

cos (3 x) cos (x) - 2 cos (2 x) cos (x)

使用倍角公式:

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

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

cos (3 x) cos (x) - (- 1 + 2 cos^{2} (x)) \cdot 2 cos (x) = 0

分解

cos (3 x) cos (x) - (- 1 + 2 cos^{2} (x)) \cdot 2 cos (x) : cos (x) (cos (3 x) - 2 (2 cos (x) + 1) (2 cos (x) - 1))

cos (3 x) cos (x) - (- 1 + 2 cos^{2} (x)) \cdot 2 cos (x)

因式分解出通项

cos (x)

= cos (x) (cos (3 x) - 2 (- 1 + cos^{2} (x) \cdot 2))

分解

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

cos (3 x) - 2 (- 1 + cos^{2} (x) \cdot 2)

分解

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

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

将

2 cos^{2} (x) - 1

改写为

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

2 cos^{2} (x) - 1

使用根式运算法则:

a = (a)^{2}

2 = (2)^{2}

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

将

1

改写为

1^{2}

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

使用指数法则:

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

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

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

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

使用平方差公式:

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

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

= (2 cos (x) + 1) (2 cos (x) - 1)

= cos (3 x) - 2 (2 cos (x) + 1) (2 cos (x) - 1)

= cos (x) (cos (3 x) - 2 (2 cos (x) + 1) (2 cos (x) - 1))

cos (x) (cos (3 x) - 2 (2 cos (x) + 1) (2 cos (x) - 1)) = 0

分别求解每个部分

cos (x) = 0 or cos (3 x) - 2 (2 cos (x) + 1) (2 cos (x) - 1) = 0

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

cos (x) = 0

的通解

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} 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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (3 x) - 2 (2 cos (x) + 1) (2 cos (x) - 1) = 0 : x = arccos (0.50000 \dots) + 2 πn, x = 2 π - arccos (0.50000 \dots) + 2 πn, x = arccos (- 0.78077 \dots) + 2 πn, x = - arccos (- 0.78077 \dots) + 2 πn

cos (3 x) - 2 (2 cos (x) + 1) (2 cos (x) - 1) = 0

使用三角恒等式改写

cos (3 x) - (- 1 + cos (x) 2) (1 + cos (x) 2) \cdot 2

cos (3 x) = 4 cos^{3} (x) - 3 cos (x)

cos (3 x)

使用三角恒等式改写

cos (3 x)

改写为

= cos (2 x + x)

使用角和恒等式:

cos (s + t) = cos (s) cos (t) - sin (s) sin (t)

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

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

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

化简

cos (2 x) cos (x) - 2 sin (x) cos (x) sin (x) : cos (x) cos (2 x) - 2 sin^{2} (x) cos (x)

cos (2 x) cos (x) - 2 sin (x) cos (x) sin (x)

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

2 sin (x) cos (x) sin (x)

使用指数法则:

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

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

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

数字相加:

1 + 1 = 2

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

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

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

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

使用倍角公式:

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

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

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

乘开

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

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

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

乘开

cos (x) (2 cos^{2} (x) - 1) : 2 cos^{3} (x) - cos (x)

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

使用分配律:

a (b - c) = ab - a c

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

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

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

化简

2 cos^{2} (x) cos (x) - 1 \cdot cos (x) : 2 cos^{3} (x) - cos (x)

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

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

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

使用指数法则:

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

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

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

数字相加:

2 + 1 = 3

= 2 cos^{3} (x)

1 \cdot cos (x) = cos (x)

1 cos (x)

乘以:

1 \cdot cos (x) = cos (x)

= cos (x)

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

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

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

乘开

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

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

使用分配律:

a (b - c) = ab - a c

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

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

使用加减运算法则

- (- a) = a

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

化简

- 2 \cdot 1 \cdot cos (x) + 2 cos^{2} (x) cos (x) : - 2 cos (x) + 2 cos^{3} (x)

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

2 \cdot 1 \cdot cos (x) = 2 cos (x)

2 \cdot 1 cos (x)

数字相乘:

2 \cdot 1 = 2

= 2 cos (x)

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

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

使用指数法则:

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

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

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

数字相加:

2 + 1 = 3

= 2 cos^{3} (x)

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

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

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

化简

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

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

对同类项分组

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

同类项相加:

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

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

同类项相加:

- cos (x) - 2 cos (x) = - 3 cos (x)

= 4 cos^{3} (x) - 3 cos (x)

= 4 cos^{3} (x) - 3 cos (x)

= 4 cos^{3} (x) - 3 cos (x)

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

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

用替代法求解

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

令

: cos (x) = u

- 3 u + 4 u^{3} - (- 1 + u 2) (1 + u 2) \cdot 2 = 0

- 3 u + 4 u^{3} - (- 1 + u 2) (1 + u 2) \cdot 2 = 0 : u \approx 1.28077 \dots, u \approx 0.50000 \dots, u \approx - 0.78077 \dots

- 3 u + 4 u^{3} - (- 1 + u 2) (1 + u 2) \cdot 2 = 0

展开

- 3 u + 4 u^{3} - (- 1 + u 2) (1 + u 2) \cdot 2 : - 3 u + 4 u^{3} - 4 u^{2} + 2

- 3 u + 4 u^{3} - (- 1 + u 2) (1 + u 2) \cdot 2

= - 3 u + 4 u^{3} - 2 (- 1 + 2 u) (1 + 2 u)

乘开

- (- 1 + u 2) (1 + u 2) \cdot 2 : - 4 u^{2} + 2

乘开

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

(- 1 + u 2) (1 + u 2)

使用平方差公式:

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

a = u 2, b = 1

= (u 2)^{2} - 1^{2}

化简

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

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

使用法则

1^{a} = 1

1^{2} = 1

= (2 u)^{2} - 1

(u 2)^{2} = 2 u^{2}

(u 2)^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

= (2)^{2} u^{2}

(2)^{2} : 2

使用根式运算法则:

a = a^{\frac{1}{2}}

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

使用指数法则:

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

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

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

\frac{1}{2} \cdot 2

分式相乘:

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

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

约分:

2

= 1

= 2

= u^{2} \cdot 2

= 2 u^{2} - 1

= 2 u^{2} - 1

= - 2 (2 u^{2} - 1)

乘开

- 2 (2 u^{2} - 1) : - 4 u^{2} + 2

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

使用分配律:

a (b - c) = ab - a c

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

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

使用加减运算法则

- (- a) = a

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

化简

- 2 \cdot 2 u^{2} + 2 \cdot 1 : - 4 u^{2} + 2

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

数字相乘:

2 \cdot 2 = 4

= - 4 u^{2} + 2 \cdot 1

数字相乘:

2 \cdot 1 = 2

= - 4 u^{2} + 2

= - 4 u^{2} + 2

= - 4 u^{2} + 2

= - 3 u + 4 u^{3} - 4 u^{2} + 2

- 3 u + 4 u^{3} - 4 u^{2} + 2 = 0

改写成标准形式

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

4 u^{3} - 4 u^{2} - 3 u + 2 = 0

使用牛顿-拉弗森方法找到

4 u^{3} - 4 u^{2} - 3 u + 2 = 0

的一个解:

u \approx 1.28077 \dots

4 u^{3} - 4 u^{2} - 3 u + 2 = 0

牛顿-拉弗森近似法定义

f (u) = 4 u^{3} - 4 u^{2} - 3 u + 2

找到

f^{^{'}} (u) : 12 u^{2} - 8 u - 3

\frac{d}{d u} (4 u^{3} - 4 u^{2} - 3 u + 2)

使用微分加减法定则:

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

= \frac{d}{d u} (4 u^{3}) - \frac{d}{d u} (4 u^{2}) - \frac{d}{d u} (3 u) + \frac{d}{d u} (2)

\frac{d}{d u} (4 u^{3}) = 12 u^{2}

\frac{d}{d u} (4 u^{3})

将常数提出:

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

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

使用幂法则:

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

= 4 \cdot 3 u^{3 - 1}

化简

= 12 u^{2}

\frac{d}{d u} (4 u^{2}) = 8 u

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

将常数提出:

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

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

使用幂法则:

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

= 4 \cdot 2 u^{2 - 1}

化简

= 8 u

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

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

将常数提出:

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

= 3 \frac{d u}{d u}

使用常见微分定则:

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

= 3 \cdot 1

化简

= 3

\frac{d}{d u} (2) = 0

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

常数微分:

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

= 0

= 12 u^{2} - 8 u - 3 + 0

化简

= 12 u^{2} - 8 u - 3

令

u_{0} = 1

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 2 : Δ u_{1} = 1

f (u_{0}) = 4 \cdot 1^{3} - 4 \cdot 1^{2} - 3 \cdot 1 + 2 = - 1

f^{^{'}} (u_{0}) = 12 \cdot 1^{2} - 8 \cdot 1 - 3 = 1

u_{1} = 2

Δ u_{1} = ∣ 2 - 1 ∣ = 1

Δ u_{1} = 1

u_{2} = 1.58620 \dots : Δ u_{2} = 0.41379 \dots

f (u_{1}) = 4 \cdot 2^{3} - 4 \cdot 2^{2} - 3 \cdot 2 + 2 = 12

f^{^{'}} (u_{1}) = 12 \cdot 2^{2} - 8 \cdot 2 - 3 = 29

u_{2} = 1.58620 \dots

Δ u_{2} = ∣ 1.58620 \dots - 2 ∣ = 0.41379 \dots

Δ u_{2} = 0.41379 \dots

u_{3} = 1.36962 \dots : Δ u_{3} = 0.21658 \dots

f (u_{2}) = 4 \cdot 1.58620 \dots^{3} - 4 \cdot 1.58620 \dots^{2} - 3 \cdot 1.58620 \dots + 2 = 3.14108 \dots

f^{^{'}} (u_{2}) = 12 \cdot 1.58620 \dots^{2} - 8 \cdot 1.58620 \dots - 3 = 14.50297 \dots

u_{3} = 1.36962 \dots

Δ u_{3} = ∣ 1.36962 \dots - 1.58620 \dots ∣ = 0.21658 \dots

Δ u_{3} = 0.21658 \dots

u_{4} = 1.29192 \dots : Δ u_{4} = 0.07769 \dots

f (u_{3}) = 4 \cdot 1.36962 \dots^{3} - 4 \cdot 1.36962 \dots^{2} - 3 \cdot 1.36962 \dots + 2 = 0.66459 \dots

f^{^{'}} (u_{3}) = 12 \cdot 1.36962 \dots^{2} - 8 \cdot 1.36962 \dots - 3 = 8.55345 \dots

u_{4} = 1.29192 \dots

Δ u_{4} = ∣ 1.29192 \dots - 1.36962 \dots ∣ = 0.07769 \dots

Δ u_{4} = 0.07769 \dots

u_{5} = 1.28098 \dots : Δ u_{5} = 0.01093 \dots

f (u_{4}) = 4 \cdot 1.29192 \dots^{3} - 4 \cdot 1.29192 \dots^{2} - 3 \cdot 1.29192 \dots + 2 = 0.07319 \dots

f^{^{'}} (u_{4}) = 12 \cdot 1.29192 \dots^{2} - 8 \cdot 1.29192 \dots - 3 = 6.69344 \dots

u_{5} = 1.28098 \dots

Δ u_{5} = ∣ 1.28098 \dots - 1.29192 \dots ∣ = 0.01093 \dots

Δ u_{5} = 0.01093 \dots

u_{6} = 1.28077 \dots : Δ u_{6} = 0.00021 \dots

f (u_{5}) = 4 \cdot 1.28098 \dots^{3} - 4 \cdot 1.28098 \dots^{2} - 3 \cdot 1.28098 \dots + 2 = 0.00137 \dots

f^{^{'}} (u_{5}) = 12 \cdot 1.28098 \dots^{2} - 8 \cdot 1.28098 \dots - 3 = 6.44328 \dots

u_{6} = 1.28077 \dots

Δ u_{6} = ∣ 1.28077 \dots - 1.28098 \dots ∣ = 0.00021 \dots

Δ u_{6} = 0.00021 \dots

u_{7} = 1.28077 \dots : Δ u_{7} = 7.99005 E - 8

f (u_{6}) = 4 \cdot 1.28077 \dots^{3} - 4 \cdot 1.28077 \dots^{2} - 3 \cdot 1.28077 \dots + 2 = 5.14435 E - 7

f^{^{'}} (u_{6}) = 12 \cdot 1.28077 \dots^{2} - 8 \cdot 1.28077 \dots - 3 = 6.43844 \dots

u_{7} = 1.28077 \dots

Δ u_{7} = ∣ 1.28077 \dots - 1.28077 \dots ∣ = 7.99005 E - 8

Δ u_{7} = 7.99005 E - 8

u \approx 1.28077 \dots

使用长除法

Eq u a t i o n 0 : \frac{4 u ^{3} - 4 u ^{2} - 3 u + 2}{u - 1.28077 \dots} = 4 u^{2} + 1.12310 \dots u - 1.56155 \dots

4 u^{2} + 1.12310 \dots u - 1.56155 \dots \approx 0

使用牛顿-拉弗森方法找到

4 u^{2} + 1.12310 \dots u - 1.56155 \dots = 0

的一个解:

u \approx 0.50000 \dots

4 u^{2} + 1.12310 \dots u - 1.56155 \dots = 0

牛顿-拉弗森近似法定义

f (u) = 4 u^{2} + 1.12310 \dots u - 1.56155 \dots

找到

f^{^{'}} (u) : 8 u + 1.12310 \dots

\frac{d}{d u} (4 u^{2} + 1.12310 \dots u - 1.56155 \dots)

使用微分加减法定则:

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

= \frac{d}{d u} (4 u^{2}) + \frac{d}{d u} (1.12310 \dots u) - \frac{d}{d u} (1.56155 \dots)

\frac{d}{d u} (4 u^{2}) = 8 u

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

将常数提出:

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

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

使用幂法则:

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

= 4 \cdot 2 u^{2 - 1}

化简

= 8 u

\frac{d}{d u} (1.12310 \dots u) = 1.12310 \dots

\frac{d}{d u} (1.12310 \dots u)

将常数提出:

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

= 1.12310 \dots \frac{d u}{d u}

使用常见微分定则:

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

= 1.12310 \dots \cdot 1

化简

= 1.12310 \dots

\frac{d}{d u} (1.56155 \dots) = 0

\frac{d}{d u} (1.56155 \dots)

常数微分:

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

= 0

= 8 u + 1.12310 \dots - 0

化简

= 8 u + 1.12310 \dots

令

u_{0} = 1

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 0.60961 \dots : Δ u_{1} = 0.39038 \dots

f (u_{0}) = 4 \cdot 1^{2} + 1.12310 \dots \cdot 1 - 1.56155 \dots = 3.56155 \dots

f^{^{'}} (u_{0}) = 8 \cdot 1 + 1.12310 \dots = 9.12310 \dots

u_{1} = 0.60961 \dots

Δ u_{1} = ∣ 0.60961 \dots - 1 ∣ = 0.39038 \dots

Δ u_{1} = 0.39038 \dots

u_{2} = 0.50800 \dots : Δ u_{2} = 0.10160 \dots

f (u_{1}) = 4 \cdot 0.60961 \dots^{2} + 1.12310 \dots \cdot 0.60961 \dots - 1.56155 \dots = 0.60961 \dots

f^{^{'}} (u_{1}) = 8 \cdot 0.60961 \dots + 1.12310 \dots = 6

u_{2} = 0.50800 \dots

Δ u_{2} = ∣ 0.50800 \dots - 0.60961 \dots ∣ = 0.10160 \dots

Δ u_{2} = 0.10160 \dots

u_{3} = 0.50004 \dots : Δ u_{3} = 0.00796 \dots

f (u_{2}) = 4 \cdot 0.50800 \dots^{2} + 1.12310 \dots \cdot 0.50800 \dots - 1.56155 \dots = 0.04129 \dots

f^{^{'}} (u_{2}) = 8 \cdot 0.50800 \dots + 1.12310 \dots = 5.18718 \dots

u_{3} = 0.50004 \dots

Δ u_{3} = ∣ 0.50004 \dots - 0.50800 \dots ∣ = 0.00796 \dots

Δ u_{3} = 0.00796 \dots

u_{4} = 0.50000 \dots : Δ u_{4} = 0.00004 \dots

f (u_{3}) = 4 \cdot 0.50004 \dots^{2} + 1.12310 \dots \cdot 0.50004 \dots - 1.56155 \dots = 0.00025 \dots

f^{^{'}} (u_{3}) = 8 \cdot 0.50004 \dots + 1.12310 \dots = 5.12350 \dots

u_{4} = 0.50000 \dots

Δ u_{4} = ∣ 0.50000 \dots - 0.50004 \dots ∣ = 0.00004 \dots

Δ u_{4} = 0.00004 \dots

u_{5} = 0.5 : Δ u_{5} = 1.91092 E - 9

f (u_{4}) = 4 \cdot 0.50000 \dots^{2} + 1.12310 \dots \cdot 0.50000 \dots - 1.56155 \dots = 9.78986 E - 9

f^{^{'}} (u_{4}) = 8 \cdot 0.50000 \dots + 1.12310 \dots = 5.12310 \dots

u_{5} = 0.5

Δ u_{5} = ∣ 0.5 - 0.50000 \dots ∣ = 1.91092 E - 9

Δ u_{5} = 1.91092 E - 9

u \approx 0.50000 \dots

使用长除法

Eq u a t i o n 0 : \frac{4 u ^{2} + 1.12310 \dots u - 1.56155 \dots}{u - 0.5} = 4 u + 3.12310 \dots

4 u + 3.12310 \dots \approx 0

u \approx - 0.78077 \dots

解为

u \approx 1.28077 \dots, u \approx 0.50000 \dots, u \approx - 0.78077 \dots

u = cos (x)

代回

cos (x) \approx 1.28077 \dots, cos (x) \approx 0.50000 \dots, cos (x) \approx - 0.78077 \dots

cos (x) \approx 1.28077 \dots, cos (x) \approx 0.50000 \dots, cos (x) \approx - 0.78077 \dots

cos (x) = 1.28077 \dots :

无解

cos (x) = 1.28077 \dots

- 1 \leq cos (x) \leq 1

无解

cos (x) = 0.50000 \dots : x = arccos (0.50000 \dots) + 2 πn, x = 2 π - arccos (0.50000 \dots) + 2 πn

cos (x) = 0.50000 \dots

使用反三角函数性质

cos (x) = 0.50000 \dots

cos (x) = 0.50000 \dots

的通解

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

x = arccos (0.50000 \dots) + 2 πn, x = 2 π - arccos (0.50000 \dots) + 2 πn

x = arccos (0.50000 \dots) + 2 πn, x = 2 π - arccos (0.50000 \dots) + 2 πn

cos (x) = - 0.78077 \dots : x = arccos (- 0.78077 \dots) + 2 πn, x = - arccos (- 0.78077 \dots) + 2 πn

cos (x) = - 0.78077 \dots

使用反三角函数性质

cos (x) = - 0.78077 \dots

cos (x) = - 0.78077 \dots

的通解

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

x = arccos (- 0.78077 \dots) + 2 πn, x = - arccos (- 0.78077 \dots) + 2 πn

x = arccos (- 0.78077 \dots) + 2 πn, x = - arccos (- 0.78077 \dots) + 2 πn

合并所有解

x = arccos (0.50000 \dots) + 2 πn, x = 2 π - arccos (0.50000 \dots) + 2 πn, x = arccos (- 0.78077 \dots) + 2 πn, x = - arccos (- 0.78077 \dots) + 2 πn

合并所有解

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = arccos (0.50000 \dots) + 2 πn, x = 2 π - arccos (0.50000 \dots) + 2 πn, x = arccos (- 0.78077 \dots) + 2 πn, x = - arccos (- 0.78077 \dots) + 2 πn

以小数形式表示解

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = 1.04719 \dots + 2 πn, x = 2 π - 1.04719 \dots + 2 πn, x = 2.46670 \dots + 2 πn, x = - 2.46670 \dots + 2 πn

作图

流行的例子

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

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

cos (θ) = 0.6015

sin (θ) = \frac{- 4}{5}

solvefor θ,x=5sec(θ)

so l v e f or θ, x = 5 sec (θ)

sec(3x)-cos(30)=0,(x+35)/5

sec (3 x) - cos (3 0^{\circ}) = 0, \frac{x + 35}{5}