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受欢迎的三角函数 >

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

解答

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

解答

x = π + 2 πn, x = - 1.97861 \dots + 2 πn

+1

度数

x = 18 0^{\circ} + 36 0^{\circ} n, x = - 113.36631 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

两边加上

tan (x)

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

两边进行平方

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

两边减去

tan^{2} (x)

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

使用三角恒等式改写

(1 + sec^{2} (x) + 2 sec (x))^{2} - tan^{2} (x)

使用毕达哥拉斯恒等式:

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

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

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

化简

(1 + sec^{2} (x) + 2 sec (x))^{2} - (sec^{2} (x) - 1) : sec^{4} (x) + 4 sec^{3} (x) + 5 sec^{2} (x) + 4 sec (x) + 2

(1 + sec^{2} (x) + 2 sec (x))^{2} - (sec^{2} (x) - 1)

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

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

乘开

(1 + sec^{2} (x) + 2 sec (x)) (1 + sec^{2} (x) + 2 sec (x)) : 6 sec^{2} (x) + 4 sec (x) + sec^{4} (x) + 4 sec^{3} (x) + 1

(1 + sec^{2} (x) + 2 sec (x)) (1 + sec^{2} (x) + 2 sec (x))

打开括号

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

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

化简

1 \cdot 1 + 1 \cdot sec^{2} (x) + 1 \cdot 2 sec (x) + 1 \cdot sec^{2} (x) + sec^{2} (x) sec^{2} (x) + 2 sec^{2} (x) sec (x) + 2 \cdot 1 \cdot sec (x) + 2 sec^{2} (x) sec (x) + 2 \cdot 2 sec (x) sec (x) : 6 sec^{2} (x) + 4 sec (x) + sec^{4} (x) + 4 sec^{3} (x) + 1

1 \cdot 1 + 1 \cdot sec^{2} (x) + 1 \cdot 2 sec (x) + 1 \cdot sec^{2} (x) + sec^{2} (x) sec^{2} (x) + 2 sec^{2} (x) sec (x) + 2 \cdot 1 \cdot sec (x) + 2 sec^{2} (x) sec (x) + 2 \cdot 2 sec (x) sec (x)

对同类项分组

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

同类项相加:

2 sec^{2} (x) sec (x) + 2 sec^{2} (x) sec (x) = 4 sec^{2} (x) sec (x)

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

同类项相加:

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

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

同类项相加:

1 \cdot 2 sec (x) + 2 \cdot 1 \cdot sec (x) = 2 \cdot 2 \cdot 1 \cdot sec (x)

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

2 \cdot 2 \cdot 1 \cdot sec (x) = 4 sec (x)

2 \cdot 2 \cdot 1 \cdot sec (x)

数字相乘:

2 \cdot 2 \cdot 1 = 4

= 4 sec (x)

sec^{2} (x) sec^{2} (x) = sec^{4} (x)

sec^{2} (x) sec^{2} (x)

使用指数法则:

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

sec^{2} (x) sec^{2} (x) = sec^{2 + 2} (x)

= sec^{2 + 2} (x)

数字相加:

2 + 2 = 4

= sec^{4} (x)

4 sec^{2} (x) sec (x) = 4 sec^{3} (x)

4 sec^{2} (x) sec (x)

使用指数法则:

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

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

= 4 sec^{2 + 1} (x)

数字相加:

2 + 1 = 3

= 4 sec^{3} (x)

2 \cdot 2 sec (x) sec (x) = 4 sec^{2} (x)

2 \cdot 2 sec (x) sec (x)

数字相乘:

2 \cdot 2 = 4

= 4 sec (x) sec (x)

使用指数法则:

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

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

= 4 sec^{1 + 1} (x)

数字相加:

1 + 1 = 2

= 4 sec^{2} (x)

1 \cdot 1 = 1

1 \cdot 1

数字相乘:

1 \cdot 1 = 1

= 1

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

同类项相加:

2 sec^{2} (x) + 4 sec^{2} (x) = 6 sec^{2} (x)

= 6 sec^{2} (x) + 4 sec (x) + sec^{4} (x) + 4 sec^{3} (x) + 1

= 6 sec^{2} (x) + 4 sec (x) + sec^{4} (x) + 4 sec^{3} (x) + 1

= 6 sec^{2} (x) + 4 sec (x) + sec^{4} (x) + 4 sec^{3} (x) + 1 - (sec^{2} (x) - 1)

- (sec^{2} (x) - 1) : - sec^{2} (x) + 1

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

打开括号

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

使用加减运算法则

- (- a) = a, - (a) = - a

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

= 6 sec^{2} (x) + 4 sec (x) + sec^{4} (x) + 4 sec^{3} (x) + 1 - sec^{2} (x) + 1

化简

6 sec^{2} (x) + 4 sec (x) + sec^{4} (x) + 4 sec^{3} (x) + 1 - sec^{2} (x) + 1 : sec^{4} (x) + 4 sec^{3} (x) + 5 sec^{2} (x) + 4 sec (x) + 2

6 sec^{2} (x) + 4 sec (x) + sec^{4} (x) + 4 sec^{3} (x) + 1 - sec^{2} (x) + 1

对同类项分组

= 6 sec^{2} (x) + 4 sec (x) + sec^{4} (x) + 4 sec^{3} (x) - sec^{2} (x) + 1 + 1

同类项相加:

6 sec^{2} (x) - sec^{2} (x) = 5 sec^{2} (x)

= 5 sec^{2} (x) + 4 sec (x) + sec^{4} (x) + 4 sec^{3} (x) + 1 + 1

数字相加:

1 + 1 = 2

= sec^{4} (x) + 4 sec^{3} (x) + 5 sec^{2} (x) + 4 sec (x) + 2

= sec^{4} (x) + 4 sec^{3} (x) + 5 sec^{2} (x) + 4 sec (x) + 2

= sec^{4} (x) + 4 sec^{3} (x) + 5 sec^{2} (x) + 4 sec (x) + 2

2 + sec^{4} (x) + 4 sec (x) + 4 sec^{3} (x) + 5 sec^{2} (x) = 0

用替代法求解

2 + sec^{4} (x) + 4 sec (x) + 4 sec^{3} (x) + 5 sec^{2} (x) = 0

令

: sec (x) = u

2 + u^{4} + 4 u + 4 u^{3} + 5 u^{2} = 0

2 + u^{4} + 4 u + 4 u^{3} + 5 u^{2} = 0 : u = - 1, u \approx - 2.52137 \dots

2 + u^{4} + 4 u + 4 u^{3} + 5 u^{2} = 0

改写成标准形式

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

u^{4} + 4 u^{3} + 5 u^{2} + 4 u + 2 = 0

因式分解

u^{4} + 4 u^{3} + 5 u^{2} + 4 u + 2 : (u + 1) (u^{3} + 3 u^{2} + 2 u + 2)

u^{4} + 4 u^{3} + 5 u^{2} + 4 u + 2

使用有理根定理

a_{0} = 2, a_{n} = 1

a_{0}

的除数

: 1, 2, a_{n}

的除数

: 1

因此，检验以下有理数:

\pm \frac{1 , 2}{1}

- \frac{1}{1}

是表达式的根，所以因式分解

u + 1

= (u + 1) \frac{u ^{4} + 4 u ^{3} + 5 u ^{2} + 4 u + 2}{u + 1}

\frac{u ^{4} + 4 u ^{3} + 5 u ^{2} + 4 u + 2}{u + 1} = u^{3} + 3 u^{2} + 2 u + 2

\frac{u ^{4} + 4 u ^{3} + 5 u ^{2} + 4 u + 2}{u + 1}

对

\frac{u ^{4} + 4 u ^{3} + 5 u ^{2} + 4 u + 2}{u + 1}

做除法:

\frac{u ^{4} + 4 u ^{3} + 5 u ^{2} + 4 u + 2}{u + 1} = u^{3} + \frac{3 u ^{3} + 5 u ^{2} + 4 u + 2}{u + 1}

将分子

u^{4} + 4 u^{3} + 5 u^{2} + 4 u + 2

与除数

u + 1

的首项系数相除：

\frac{u ^{4}}{u} = u^{3}

商 = u^{3}

将

u + 1

乘以

u^{3} : u^{4} + u^{3}

将

u^{4} + 4 u^{3} + 5 u^{2} + 4 u + 2

减去

u^{4} + u^{3}

得到新的余数

余数 = 3 u^{3} + 5 u^{2} + 4 u + 2

因此

\frac{u ^{4} + 4 u ^{3} + 5 u ^{2} + 4 u + 2}{u + 1} = u^{3} + \frac{3 u ^{3} + 5 u ^{2} + 4 u + 2}{u + 1}

= u^{3} + \frac{3 u ^{3} + 5 u ^{2} + 4 u + 2}{u + 1}

对

\frac{3 u ^{3} + 5 u ^{2} + 4 u + 2}{u + 1}

做除法:

\frac{3 u ^{3} + 5 u ^{2} + 4 u + 2}{u + 1} = 3 u^{2} + \frac{2 u ^{2} + 4 u + 2}{u + 1}

将分子

3 u^{3} + 5 u^{2} + 4 u + 2

与除数

u + 1

的首项系数相除：

\frac{3 u ^{3}}{u} = 3 u^{2}

商 = 3 u^{2}

将

u + 1

乘以

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

将

3 u^{3} + 5 u^{2} + 4 u + 2

减去

3 u^{3} + 3 u^{2}

得到新的余数

余数 = 2 u^{2} + 4 u + 2

因此

\frac{3 u ^{3} + 5 u ^{2} + 4 u + 2}{u + 1} = 3 u^{2} + \frac{2 u ^{2} + 4 u + 2}{u + 1}

= u^{3} + 3 u^{2} + \frac{2 u ^{2} + 4 u + 2}{u + 1}

对

\frac{2 u ^{2} + 4 u + 2}{u + 1}

做除法:

\frac{2 u ^{2} + 4 u + 2}{u + 1} = 2 u + \frac{2 u + 2}{u + 1}

将分子

2 u^{2} + 4 u + 2

与除数

u + 1

的首项系数相除：

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

商 = 2 u

将

u + 1

乘以

2 u : 2 u^{2} + 2 u

将

2 u^{2} + 4 u + 2

减去

2 u^{2} + 2 u

得到新的余数

余数 = 2 u + 2

因此

\frac{2 u ^{2} + 4 u + 2}{u + 1} = 2 u + \frac{2 u + 2}{u + 1}

= u^{3} + 3 u^{2} + 2 u + \frac{2 u + 2}{u + 1}

对

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

做除法:

\frac{2 u + 2}{u + 1} = 2

将分子

2 u + 2

与除数

u + 1

的首项系数相除：

\frac{2 u}{u} = 2

商 = 2

将

u + 1

乘以

2 : 2 u + 2

将

2 u + 2

减去

2 u + 2

得到新的余数

余数 = 0

因此

\frac{2 u + 2}{u + 1} = 2

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

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

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

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

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

解

u + 1 = 0 : u = - 1

u + 1 = 0

将

1

到右边

u + 1 = 0

两边减去

1

u + 1 - 1 = 0 - 1

化简

u = - 1

u = - 1

解

u^{3} + 3 u^{2} + 2 u + 2 = 0 : u \approx - 2.52137 \dots

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

使用牛顿-拉弗森方法找到

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

的一个解:

u \approx - 2.52137 \dots

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

牛顿-拉弗森近似法定义

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

找到

f^{^{'}} (u) : 3 u^{2} + 6 u + 2

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

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

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

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

将常数提出:

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

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

使用幂法则:

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

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

化简

= 6 u

\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} (2) = 0

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

常数微分:

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

= 0

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

化简

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

令

u_{0} = - 2

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = - 3 : Δ u_{1} = 1

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

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

u_{1} = - 3

Δ u_{1} = ∣ - 3 - (- 2) ∣ = 1

Δ u_{1} = 1

u_{2} = - 2.63636 \dots : Δ u_{2} = 0.36363 \dots

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

f^{^{'}} (u_{1}) = 3 (- 3)^{2} + 6 (- 3) + 2 = 11

u_{2} = - 2.63636 \dots

Δ u_{2} = ∣ - 2.63636 \dots - (- 3) ∣ = 0.36363 \dots

Δ u_{2} = 0.36363 \dots

u_{3} = - 2.53039 \dots : Δ u_{3} = 0.10597 \dots

f (u_{2}) = (- 2.63636 \dots)^{3} + 3 (- 2.63636 \dots)^{2} + 2 (- 2.63636 \dots) + 2 = - 0.74530 \dots

f^{^{'}} (u_{2}) = 3 (- 2.63636 \dots)^{2} + 6 (- 2.63636 \dots) + 2 = 7.03305 \dots

u_{3} = - 2.53039 \dots

Δ u_{3} = ∣ - 2.53039 \dots - (- 2.63636 \dots) ∣ = 0.10597 \dots

Δ u_{3} = 0.10597 \dots

u_{4} = - 2.52144 \dots : Δ u_{4} = 0.00895 \dots

f (u_{3}) = (- 2.53039 \dots)^{3} + 3 (- 2.53039 \dots)^{2} + 2 (- 2.53039 \dots) + 2 = - 0.05393 \dots

f^{^{'}} (u_{3}) = 3 (- 2.53039 \dots)^{2} + 6 (- 2.53039 \dots) + 2 = 6.02629 \dots

u_{4} = - 2.52144 \dots

Δ u_{4} = ∣ - 2.52144 \dots - (- 2.53039 \dots) ∣ = 0.00895 \dots

Δ u_{4} = 0.00895 \dots

u_{5} = - 2.52137 \dots : Δ u_{5} = 0.00006 \dots

f (u_{4}) = (- 2.52144 \dots)^{3} + 3 (- 2.52144 \dots)^{2} + 2 (- 2.52144 \dots) + 2 = - 0.00036 \dots

f^{^{'}} (u_{4}) = 3 (- 2.52144 \dots)^{2} + 6 (- 2.52144 \dots) + 2 = 5.94435 \dots

u_{5} = - 2.52137 \dots

Δ u_{5} = ∣ - 2.52137 \dots - (- 2.52144 \dots) ∣ = 0.00006 \dots

Δ u_{5} = 0.00006 \dots

u_{6} = - 2.52137 \dots : Δ u_{6} = 2.92858 E - 9

f (u_{5}) = (- 2.52137 \dots)^{3} + 3 (- 2.52137 \dots)^{2} + 2 (- 2.52137 \dots) + 2 = - 1.74069 E - 8

f^{^{'}} (u_{5}) = 3 (- 2.52137 \dots)^{2} + 6 (- 2.52137 \dots) + 2 = 5.94378 \dots

u_{6} = - 2.52137 \dots

Δ u_{6} = ∣ - 2.52137 \dots - (- 2.52137 \dots) ∣ = 2.92858 E - 9

Δ u_{6} = 2.92858 E - 9

u \approx - 2.52137 \dots

使用长除法

Eq u a t i o n 0 : \frac{u ^{3} + 3 u ^{2} + 2 u + 2}{u + 2.52137 \dots} = u^{2} + 0.47862 \dots u + 0.79321 \dots

u^{2} + 0.47862 \dots u + 0.79321 \dots \approx 0

使用牛顿-拉弗森方法找到

u^{2} + 0.47862 \dots u + 0.79321 \dots = 0

的一个解:

u \in R

无解

u^{2} + 0.47862 \dots u + 0.79321 \dots = 0

牛顿-拉弗森近似法定义

f (u) = u^{2} + 0.47862 \dots u + 0.79321 \dots

找到

f^{^{'}} (u) : 2 u + 0.47862 \dots

\frac{d}{d u} (u^{2} + 0.47862 \dots u + 0.79321 \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} (0.47862 \dots u) + \frac{d}{d u} (0.79321 \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} (0.47862 \dots u) = 0.47862 \dots

\frac{d}{d u} (0.47862 \dots u)

将常数提出:

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

= 0.47862 \dots \frac{d u}{d u}

使用常见微分定则:

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

= 0.47862 \dots \cdot 1

化简

= 0.47862 \dots

\frac{d}{d u} (0.79321 \dots) = 0

\frac{d}{d u} (0.79321 \dots)

常数微分:

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

= 0

= 2 u + 0.47862 \dots + 0

化简

= 2 u + 0.47862 \dots

令

u_{0} = - 2

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = - 0.91066 \dots : Δ u_{1} = 1.08933 \dots

f (u_{0}) = (- 2)^{2} + 0.47862 \dots (- 2) + 0.79321 \dots = 3.83597 \dots

f^{^{'}} (u_{0}) = 2 (- 2) + 0.47862 \dots = - 3.52137 \dots

u_{1} = - 0.91066 \dots

Δ u_{1} = ∣ - 0.91066 \dots - (- 2) ∣ = 1.08933 \dots

Δ u_{1} = 1.08933 \dots

u_{2} = - 0.02687 \dots : Δ u_{2} = 0.88378 \dots

f (u_{1}) = (- 0.91066 \dots)^{2} + 0.47862 \dots (- 0.91066 \dots) + 0.79321 \dots = 1.18665 \dots

f^{^{'}} (u_{1}) = 2 (- 0.91066 \dots) + 0.47862 \dots = - 1.34270 \dots

u_{2} = - 0.02687 \dots

Δ u_{2} = ∣ - 0.02687 \dots - (- 0.91066 \dots) ∣ = 0.88378 \dots

Δ u_{2} = 0.88378 \dots

u_{3} = - 1.86527 \dots : Δ u_{3} = 1.83839 \dots

f (u_{2}) = (- 0.02687 \dots)^{2} + 0.47862 \dots (- 0.02687 \dots) + 0.79321 \dots = 0.78107 \dots

f^{^{'}} (u_{2}) = 2 (- 0.02687 \dots) + 0.47862 \dots = 0.42486 \dots

u_{3} = - 1.86527 \dots

Δ u_{3} = ∣ - 1.86527 \dots - (- 0.02687 \dots) ∣ = 1.83839 \dots

Δ u_{3} = 1.83839 \dots

u_{4} = - 0.82598 \dots : Δ u_{4} = 1.03929 \dots

f (u_{3}) = (- 1.86527 \dots)^{2} + 0.47862 \dots (- 1.86527 \dots) + 0.79321 \dots = 3.37970 \dots

f^{^{'}} (u_{3}) = 2 (- 1.86527 \dots) + 0.47862 \dots = - 3.25192 \dots

u_{4} = - 0.82598 \dots

Δ u_{4} = ∣ - 0.82598 \dots - (- 1.86527 \dots) ∣ = 1.03929 \dots

Δ u_{4} = 1.03929 \dots

u_{5} = 0.09457 \dots : Δ u_{5} = 0.92055 \dots

f (u_{4}) = (- 0.82598 \dots)^{2} + 0.47862 \dots (- 0.82598 \dots) + 0.79321 \dots = 1.08013 \dots

f^{^{'}} (u_{4}) = 2 (- 0.82598 \dots) + 0.47862 \dots = - 1.17334 \dots

u_{5} = 0.09457 \dots

Δ u_{5} = ∣ 0.09457 \dots - (- 0.82598 \dots) ∣ = 0.92055 \dots

Δ u_{5} = 0.92055 \dots

u_{6} = - 1.17445 \dots : Δ u_{6} = 1.26903 \dots

f (u_{5}) = 0.09457 \dots^{2} + 0.47862 \dots \cdot 0.09457 \dots + 0.79321 \dots = 0.84742 \dots

f^{^{'}} (u_{5}) = 2 \cdot 0.09457 \dots + 0.47862 \dots = 0.66777 \dots

u_{6} = - 1.17445 \dots

Δ u_{6} = ∣ - 1.17445 \dots - 0.09457 \dots ∣ = 1.26903 \dots

Δ u_{6} = 1.26903 \dots

u_{7} = - 0.31338 \dots : Δ u_{7} = 0.86106 \dots

f (u_{6}) = (- 1.17445 \dots)^{2} + 0.47862 \dots (- 1.17445 \dots) + 0.79321 \dots = 1.61044 \dots

f^{^{'}} (u_{6}) = 2 (- 1.17445 \dots) + 0.47862 \dots = - 1.87029 \dots

u_{7} = - 0.31338 \dots

Δ u_{7} = ∣ - 0.31338 \dots - (- 1.17445 \dots) ∣ = 0.86106 \dots

Δ u_{7} = 0.86106 \dots

u_{8} = 4.69091 \dots : Δ u_{8} = 5.00430 \dots

f (u_{7}) = (- 0.31338 \dots)^{2} + 0.47862 \dots (- 0.31338 \dots) + 0.79321 \dots = 0.74143 \dots

f^{^{'}} (u_{7}) = 2 (- 0.31338 \dots) + 0.47862 \dots = - 0.14815 \dots

u_{8} = 4.69091 \dots

Δ u_{8} = ∣ 4.69091 \dots - (- 0.31338 \dots) ∣ = 5.00430 \dots

Δ u_{8} = 5.00430 \dots

u_{9} = 2.15116 \dots : Δ u_{9} = 2.53974 \dots

f (u_{8}) = 4.69091 \dots^{2} + 0.47862 \dots \cdot 4.69091 \dots + 0.79321 \dots = 25.04307 \dots

f^{^{'}} (u_{8}) = 2 \cdot 4.69091 \dots + 0.47862 \dots = 9.86045 \dots

u_{9} = 2.15116 \dots

Δ u_{9} = ∣ 2.15116 \dots - 4.69091 \dots ∣ = 2.53974 \dots

Δ u_{9} = 2.53974 \dots

u_{10} = 0.80199 \dots : Δ u_{10} = 1.34917 \dots

f (u_{9}) = 2.15116 \dots^{2} + 0.47862 \dots \cdot 2.15116 \dots + 0.79321 \dots = 6.45032 \dots

f^{^{'}} (u_{9}) = 2 \cdot 2.15116 \dots + 0.47862 \dots = 4.78095 \dots

u_{10} = 0.80199 \dots

Δ u_{10} = ∣ 0.80199 \dots - 2.15116 \dots ∣ = 1.34917 \dots

Δ u_{10} = 1.34917 \dots

u_{11} = - 0.07203 \dots : Δ u_{11} = 0.87402 \dots

f (u_{10}) = 0.80199 \dots^{2} + 0.47862 \dots \cdot 0.80199 \dots + 0.79321 \dots = 1.82026 \dots

f^{^{'}} (u_{10}) = 2 \cdot 0.80199 \dots + 0.47862 \dots = 2.08261 \dots

u_{11} = - 0.07203 \dots

Δ u_{11} = ∣ - 0.07203 \dots - 0.80199 \dots ∣ = 0.87402 \dots

Δ u_{11} = 0.87402 \dots

u_{12} = - 2.35547 \dots : Δ u_{12} = 2.28344 \dots

f (u_{11}) = (- 0.07203 \dots)^{2} + 0.47862 \dots (- 0.07203 \dots) + 0.79321 \dots = 0.76392 \dots

f^{^{'}} (u_{11}) = 2 (- 0.07203 \dots) + 0.47862 \dots = 0.33455 \dots

u_{12} = - 2.35547 \dots

Δ u_{12} = ∣ - 2.35547 \dots - (- 0.07203 \dots) ∣ = 2.28344 \dots

Δ u_{12} = 2.28344 \dots

无法得出解

解是

u \approx - 2.52137 \dots

解为

u = - 1, u \approx - 2.52137 \dots

u = sec (x)

代回

sec (x) = - 1, sec (x) \approx - 2.52137 \dots

sec (x) = - 1, sec (x) \approx - 2.52137 \dots

sec (x) = - 1 : x = π + 2 πn

sec (x) = - 1

sec (x) = - 1

的通解

sec (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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} 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} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

x = π + 2 πn

x = π + 2 πn

sec (x) = - 2.52137 \dots : x = arcsec (- 2.52137 \dots) + 2 πn, x = - arcsec (- 2.52137 \dots) + 2 πn

sec (x) = - 2.52137 \dots

使用反三角函数性质

sec (x) = - 2.52137 \dots

sec (x) = - 2.52137 \dots

的通解

sec (x) = - a \Rightarrow x = arcsec (- a) + 2 πn, x = - arcsec (- a) + 2 πn

x = arcsec (- 2.52137 \dots) + 2 πn, x = - arcsec (- 2.52137 \dots) + 2 πn

x = arcsec (- 2.52137 \dots) + 2 πn, x = - arcsec (- 2.52137 \dots) + 2 πn

合并所有解

x = π + 2 πn, x = arcsec (- 2.52137 \dots) + 2 πn, x = - arcsec (- 2.52137 \dots) + 2 πn

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

π + 2 πn

的解:

真

π + 2 πn

代入

n = 1

π + 2 π 1

对于

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

代入

x = π + 2 π 1

(sec (π + 2 π 1) + 1)^{2} - tan (π + 2 π 1) = 0

整理后得

0 = 0

\Rightarrow 真

检验

arcsec (- 2.52137 \dots) + 2 πn

的解:

假

arcsec (- 2.52137 \dots) + 2 πn

代入

n = 1

arcsec (- 2.52137 \dots) + 2 π 1

对于

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

代入

x = arcsec (- 2.52137 \dots) + 2 π 1

(sec (arcsec (- 2.52137 \dots) + 2 π 1) + 1)^{2} - tan (arcsec (- 2.52137 \dots) + 2 π 1) = 0

整理后得

4.62919 \dots = 0

\Rightarrow 假

检验

- arcsec (- 2.52137 \dots) + 2 πn

的解:

真

- arcsec (- 2.52137 \dots) + 2 πn

代入

n = 1

- arcsec (- 2.52137 \dots) + 2 π 1

对于

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

代入

x = - arcsec (- 2.52137 \dots) + 2 π 1

(sec (- arcsec (- 2.52137 \dots) + 2 π 1) + 1)^{2} - tan (- arcsec (- 2.52137 \dots) + 2 π 1) = 0

整理后得

0 = 0

\Rightarrow 真

x = π + 2 πn, x = - arcsec (- 2.52137 \dots) + 2 πn

以小数形式表示解

x = π + 2 πn, x = - 1.97861 \dots + 2 πn

作图

流行的例子

cos (A) = \frac{8}{17}

sec(θ)= 6/(-5),sec(-θ)

sec (θ) = \frac{6}{- 5}, sec (- θ)

cos (π z) = 0

sin(x)*cos(x)-3cos(x)=0

sin (x) \cdot cos (x) - 3 cos (x) = 0

9.8*sin(α)-0.2*9.8*cos(α)=0.47

9.8 \cdot sin (α) - 0.2 \cdot 9.8 \cdot cos (α) = 0.47