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受欢迎的三角函数 >

tan^2(x)= 1/(cos(x)+1)

解答

tan^{2} (x) = \frac{1}{cos ( x ) + 1}

解答

x = 0.64026 \dots + 2 πn, x = 2 π - 0.64026 \dots + 2 πn, x = 2.15910 \dots + 2 πn, x = - 2.15910 \dots + 2 πn

+1

度数

x = 36.68445 \dots^{\circ} + 36 0^{\circ} n, x = 323.31554 \dots^{\circ} + 36 0^{\circ} n, x = 123.70783 \dots^{\circ} + 36 0^{\circ} n, x = - 123.70783 \dots^{\circ} + 36 0^{\circ} n

求解步骤

tan^{2} (x) = \frac{1}{cos ( x ) + 1}

两边进行平方

(tan^{2} (x))^{2} = (\frac{1}{cos ( x ) + 1})^{2}

两边减去

(\frac{1}{cos ( x ) + 1})^{2}

tan^{4} (x) - \frac{1}{( cos ( x ) + 1 ) ^{2}} = 0

化简

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

tan^{4} (x) - \frac{1}{( cos ( x ) + 1 ) ^{2}}

将项转换为分式:

tan^{4} (x) = \frac{tan ^{4} ( x ) ( cos ( x ) + 1 ) ^{2}}{( cos ( x ) + 1 ) ^{2}}

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

因为分母相等，所以合并分式:

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

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

\frac{tan ^{4} ( x ) ( cos ( x ) + 1 ) ^{2} - 1}{( cos ( x ) + 1 ) ^{2}} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

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

使用三角恒等式改写

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

使用基本三角恒等式:

cos (x) = \frac{1}{sec ( x )}

= - 1 + (1 + \frac{1}{sec ( x )})^{2} tan^{4} (x)

- 1 + (1 + \frac{1}{sec ( x )})^{2} tan^{4} (x) = 0

分解

- 1 + (1 + \frac{1}{sec ( x )})^{2} tan^{4} (x) : (tan^{2} (x) (1 + \frac{1}{sec ( x )}) + 1) (tan^{2} (x) (1 + \frac{1}{sec ( x )}) - 1)

- 1 + (1 + \frac{1}{sec ( x )})^{2} tan^{4} (x)

将

- 1 + (1 + \frac{1}{sec ( x )})^{2} tan^{4} (x)

改写为

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

- 1 + (1 + \frac{1}{sec ( x )})^{2} tan^{4} (x)

使用指数法则:

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

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

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

使用指数法则:

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

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

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

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

使用平方差公式:

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

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

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

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

分别求解每个部分

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

tan^{2} (x) (1 + \frac{1}{sec ( x )}) + 1 = 0 :

无解

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

使用三角恒等式改写

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

化简

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

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

乘开

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

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

使用 FOIL 方法:

(a + b) (c + d) = a c + a d + b c + b d

a = 1, b = \frac{1}{sec ( x )}, c = sec^{2} (x), d = - 1

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

使用加减运算法则

+ (- a) = - a

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

化简

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

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

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

1 \cdot sec^{2} (x)

乘以:

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

= sec^{2} (x)

1 \cdot 1 = 1

1 \cdot 1

数字相乘:

1 \cdot 1 = 1

= 1

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

\frac{1}{sec ( x )} sec^{2} (x)

分式相乘:

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

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

乘以:

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

= \frac{sec ^{2} ( x )}{sec ( x )}

约分:

sec (x)

= sec (x)

1 \cdot \frac{1}{sec ( x )} = \frac{1}{sec ( x )}

1 \cdot \frac{1}{sec ( x )}

乘以:

1 \cdot \frac{1}{sec ( x )} = \frac{1}{sec ( x )}

= \frac{1}{sec ( x )}

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

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

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

化简

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

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

对同类项分组

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

1 - 1 = 0

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

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

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

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

用替代法求解

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

令

: sec (x) = u

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

- \frac{1}{u} + u + u^{2} = 0 : u \approx 0.75487 \dots

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

在两边乘以

u

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

在两边乘以

u

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

化简

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

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

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

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

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

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

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

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

解

- 1 + u^{2} + u^{3} = 0 : u \approx 0.75487 \dots

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

改写成标准形式

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

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

使用牛顿-拉弗森方法找到

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

的一个解:

u \approx 0.75487 \dots

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

牛顿-拉弗森近似法定义

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

找到

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

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

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

\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} (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

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

化简

= 3 u^{2} + 2 u

令

u_{0} = 1

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 0.8 : Δ u_{1} = 0.2

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

f^{^{'}} (u_{0}) = 3 \cdot 1^{2} + 2 \cdot 1 = 5

u_{1} = 0.8

Δ u_{1} = ∣ 0.8 - 1 ∣ = 0.2

Δ u_{1} = 0.2

u_{2} = 0.75681 \dots : Δ u_{2} = 0.04318 \dots

f (u_{1}) = 0. 8^{3} + 0. 8^{2} - 1 = 0.152

f^{^{'}} (u_{1}) = 3 \cdot 0. 8^{2} + 2 \cdot 0.8 = 3.52

u_{2} = 0.75681 \dots

Δ u_{2} = ∣ 0.75681 \dots - 0.8 ∣ = 0.04318 \dots

Δ u_{2} = 0.04318 \dots

u_{3} = 0.75488 \dots : Δ u_{3} = 0.00193 \dots

f (u_{2}) = 0.75681 \dots^{3} + 0.75681 \dots^{2} - 1 = 0.00625 \dots

f^{^{'}} (u_{2}) = 3 \cdot 0.75681 \dots^{2} + 2 \cdot 0.75681 \dots = 3.23195 \dots

u_{3} = 0.75488 \dots

Δ u_{3} = ∣ 0.75488 \dots - 0.75681 \dots ∣ = 0.00193 \dots

Δ u_{3} = 0.00193 \dots

u_{4} = 0.75487 \dots : Δ u_{4} = 3.80818 E - 6

f (u_{3}) = 0.75488 \dots^{3} + 0.75488 \dots^{2} - 1 = 0.00001 \dots

f^{^{'}} (u_{3}) = 3 \cdot 0.75488 \dots^{2} + 2 \cdot 0.75488 \dots = 3.21930 \dots

u_{4} = 0.75487 \dots

Δ u_{4} = ∣ 0.75487 \dots - 0.75488 \dots ∣ = 3.80818 E - 6

Δ u_{4} = 3.80818 E - 6

u_{5} = 0.75487 \dots : Δ u_{5} = 1.47065 E - 11

f (u_{4}) = 0.75487 \dots^{3} + 0.75487 \dots^{2} - 1 = 4.73444 E - 11

f^{^{'}} (u_{4}) = 3 \cdot 0.75487 \dots^{2} + 2 \cdot 0.75487 \dots = 3.21927 \dots

u_{5} = 0.75487 \dots

Δ u_{5} = ∣ 0.75487 \dots - 0.75487 \dots ∣ = 1.47065 E - 11

Δ u_{5} = 1.47065 E - 11

u \approx 0.75487 \dots

使用长除法

Eq u a t i o n 0 : \frac{u ^{3} + u ^{2} - 1}{u - 0.75487 \dots} = u^{2} + 1.75487 \dots u + 1.32471 \dots

u^{2} + 1.75487 \dots u + 1.32471 \dots \approx 0

使用牛顿-拉弗森方法找到

u^{2} + 1.75487 \dots u + 1.32471 \dots = 0

的一个解:

u \in R

无解

u^{2} + 1.75487 \dots u + 1.32471 \dots = 0

牛顿-拉弗森近似法定义

f (u) = u^{2} + 1.75487 \dots u + 1.32471 \dots

找到

f^{^{'}} (u) : 2 u + 1.75487 \dots

\frac{d}{d u} (u^{2} + 1.75487 \dots u + 1.32471 \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} (1.75487 \dots u) + \frac{d}{d u} (1.32471 \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} (1.75487 \dots u) = 1.75487 \dots

\frac{d}{d u} (1.75487 \dots u)

将常数提出:

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

= 1.75487 \dots \frac{d u}{d u}

使用常见微分定则:

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

= 1.75487 \dots \cdot 1

化简

= 1.75487 \dots

\frac{d}{d u} (1.32471 \dots) = 0

\frac{d}{d u} (1.32471 \dots)

常数微分:

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

= 0

= 2 u + 1.75487 \dots + 0

化简

= 2 u + 1.75487 \dots

令

u_{0} = - 1

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 1.32471 \dots : Δ u_{1} = 2.32471 \dots

f (u_{0}) = (- 1)^{2} + 1.75487 \dots (- 1) + 1.32471 \dots = 0.56984 \dots

f^{^{'}} (u_{0}) = 2 (- 1) + 1.75487 \dots = - 0.24512 \dots

u_{1} = 1.32471 \dots

Δ u_{1} = ∣ 1.32471 \dots - (- 1) ∣ = 2.32471 \dots

Δ u_{1} = 2.32471 \dots

u_{2} = 0.09766 \dots : Δ u_{2} = 1.22705 \dots

f (u_{1}) = 1.32471 \dots^{2} + 1.75487 \dots \cdot 1.32471 \dots + 1.32471 \dots = 5.40431 \dots

f^{^{'}} (u_{1}) = 2 \cdot 1.32471 \dots + 1.75487 \dots = 4.40431 \dots

u_{2} = 0.09766 \dots

Δ u_{2} = ∣ 0.09766 \dots - 1.32471 \dots ∣ = 1.22705 \dots

Δ u_{2} = 1.22705 \dots

u_{3} = - 0.67437 \dots : Δ u_{3} = 0.77204 \dots

f (u_{2}) = 0.09766 \dots^{2} + 1.75487 \dots \cdot 0.09766 \dots + 1.32471 \dots = 1.50565 \dots

f^{^{'}} (u_{2}) = 2 \cdot 0.09766 \dots + 1.75487 \dots = 1.95021 \dots

u_{3} = - 0.67437 \dots

Δ u_{3} = ∣ - 0.67437 \dots - 0.09766 \dots ∣ = 0.77204 \dots

Δ u_{3} = 0.77204 \dots

u_{4} = - 2.14204 \dots : Δ u_{4} = 1.46766 \dots

f (u_{3}) = (- 0.67437 \dots)^{2} + 1.75487 \dots (- 0.67437 \dots) + 1.32471 \dots = 0.59605 \dots

f^{^{'}} (u_{3}) = 2 (- 0.67437 \dots) + 1.75487 \dots = 0.40612 \dots

u_{4} = - 2.14204 \dots

Δ u_{4} = ∣ - 2.14204 \dots - (- 0.67437 \dots) ∣ = 1.46766 \dots

Δ u_{4} = 1.46766 \dots

u_{5} = - 1.29037 \dots : Δ u_{5} = 0.85166 \dots

f (u_{4}) = (- 2.14204 \dots)^{2} + 1.75487 \dots (- 2.14204 \dots) + 1.32471 \dots = 2.15403 \dots

f^{^{'}} (u_{4}) = 2 (- 2.14204 \dots) + 1.75487 \dots = - 2.52920 \dots

u_{5} = - 1.29037 \dots

Δ u_{5} = ∣ - 1.29037 \dots - (- 2.14204 \dots) ∣ = 0.85166 \dots

Δ u_{5} = 0.85166 \dots

u_{6} = - 0.41210 \dots : Δ u_{6} = 0.87826 \dots

f (u_{5}) = (- 1.29037 \dots)^{2} + 1.75487 \dots (- 1.29037 \dots) + 1.32471 \dots = 0.72533 \dots

f^{^{'}} (u_{5}) = 2 (- 1.29037 \dots) + 1.75487 \dots = - 0.82587 \dots

u_{6} = - 0.41210 \dots

Δ u_{6} = ∣ - 0.41210 \dots - (- 1.29037 \dots) ∣ = 0.87826 \dots

Δ u_{6} = 0.87826 \dots

u_{7} = - 1.24093 \dots : Δ u_{7} = 0.82882 \dots

f (u_{6}) = (- 0.41210 \dots)^{2} + 1.75487 \dots (- 0.41210 \dots) + 1.32471 \dots = 0.77135 \dots

f^{^{'}} (u_{6}) = 2 (- 0.41210 \dots) + 1.75487 \dots = 0.93065 \dots

u_{7} = - 1.24093 \dots

Δ u_{7} = ∣ - 1.24093 \dots - (- 0.41210 \dots) ∣ = 0.82882 \dots

Δ u_{7} = 0.82882 \dots

u_{8} = - 0.29600 \dots : Δ u_{8} = 0.94492 \dots

f (u_{7}) = (- 1.24093 \dots)^{2} + 1.75487 \dots (- 1.24093 \dots) + 1.32471 \dots = 0.68694 \dots

f^{^{'}} (u_{7}) = 2 (- 1.24093 \dots) + 1.75487 \dots = - 0.72698 \dots

u_{8} = - 0.29600 \dots

Δ u_{8} = ∣ - 0.29600 \dots - (- 1.24093 \dots) ∣ = 0.94492 \dots

Δ u_{8} = 0.94492 \dots

u_{9} = - 1.06383 \dots : Δ u_{9} = 0.76782 \dots

f (u_{8}) = (- 0.29600 \dots)^{2} + 1.75487 \dots (- 0.29600 \dots) + 1.32471 \dots = 0.89288 \dots

f^{^{'}} (u_{8}) = 2 (- 0.29600 \dots) + 1.75487 \dots = 1.16286 \dots

u_{9} = - 1.06383 \dots

Δ u_{9} = ∣ - 1.06383 \dots - (- 0.29600 \dots) ∣ = 0.76782 \dots

Δ u_{9} = 0.76782 \dots

u_{10} = 0.51763 \dots : Δ u_{10} = 1.58147 \dots

f (u_{9}) = (- 1.06383 \dots)^{2} + 1.75487 \dots (- 1.06383 \dots) + 1.32471 \dots = 0.58956 \dots

f^{^{'}} (u_{9}) = 2 (- 1.06383 \dots) + 1.75487 \dots = - 0.37279 \dots

u_{10} = 0.51763 \dots

Δ u_{10} = ∣ 0.51763 \dots - (- 1.06383 \dots) ∣ = 1.58147 \dots

Δ u_{10} = 1.58147 \dots

无法得出解

解是

u \approx 0.75487 \dots

u \approx 0.75487 \dots

验证解

找到无定义的点（奇点）:

u = 0

取

- \frac{1}{u} + u + 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 0.75487 \dots

u = sec (x)

代回

sec (x) \approx 0.75487 \dots

sec (x) \approx 0.75487 \dots

sec (x) = 0.75487 \dots :

无解

sec (x) = 0.75487 \dots

sec (x) \leq - 1 or sec (x) \geq 1

无解

合并所有解

无解

tan^{2} (x) (1 + \frac{1}{sec ( x )}) - 1 = 0 : x = arcsec (1.24697 \dots) + 2 πn, x = 2 π - arcsec (1.24697 \dots) + 2 πn, x = arcsec (- 1.80193 \dots) + 2 πn, x = - arcsec (- 1.80193 \dots) + 2 πn

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

使用三角恒等式改写

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

化简

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

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

乘开

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

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

使用 FOIL 方法:

(a + b) (c + d) = a c + a d + b c + b d

a = 1, b = \frac{1}{sec ( x )}, c = sec^{2} (x), d = - 1

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

使用加减运算法则

+ (- a) = - a

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

化简

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

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

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

1 \cdot sec^{2} (x)

乘以:

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

= sec^{2} (x)

1 \cdot 1 = 1

1 \cdot 1

数字相乘:

1 \cdot 1 = 1

= 1

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

\frac{1}{sec ( x )} sec^{2} (x)

分式相乘:

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

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

乘以:

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

= \frac{sec ^{2} ( x )}{sec ( x )}

约分:

sec (x)

= sec (x)

1 \cdot \frac{1}{sec ( x )} = \frac{1}{sec ( x )}

1 \cdot \frac{1}{sec ( x )}

乘以:

1 \cdot \frac{1}{sec ( x )} = \frac{1}{sec ( x )}

= \frac{1}{sec ( x )}

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

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

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

化简

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

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

对同类项分组

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

数字相减:

- 1 - 1 = - 2

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

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

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

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

用替代法求解

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

令

: sec (x) = u

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

- 2 - \frac{1}{u} + u + u^{2} = 0 : u \approx - 0.44504 \dots, u \approx 1.24697 \dots, u \approx - 1.80193 \dots

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

在两边乘以

u

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

在两边乘以

u

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

化简

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

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

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

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

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

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

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

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

解

- 2 u - 1 + u^{2} + u^{3} = 0 : u \approx - 0.44504 \dots, u \approx 1.24697 \dots, u \approx - 1.80193 \dots

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

改写成标准形式

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

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

使用牛顿-拉弗森方法找到

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

的一个解:

u \approx - 0.44504 \dots

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

牛顿-拉弗森近似法定义

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

找到

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

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

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

\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} (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 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} (1) = 0

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

常数微分:

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

= 0

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

化简

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

令

u_{0} = 0

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = - 0.5 : Δ u_{1} = 0.5

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

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

u_{1} = - 0.5

Δ u_{1} = ∣ - 0.5 - 0 ∣ = 0.5

Δ u_{1} = 0.5

u_{2} = - 0.44444 \dots : Δ u_{2} = 0.05555 \dots

f (u_{1}) = (- 0.5)^{3} + (- 0.5)^{2} - 2 (- 0.5) - 1 = 0.125

f^{^{'}} (u_{1}) = 3 (- 0.5)^{2} + 2 (- 0.5) - 2 = - 2.25

u_{2} = - 0.44444 \dots

Δ u_{2} = ∣ - 0.44444 \dots - (- 0.5) ∣ = 0.05555 \dots

Δ u_{2} = 0.05555 \dots

u_{3} = - 0.44504 \dots : Δ u_{3} = 0.00059 \dots

f (u_{2}) = (- 0.44444 \dots)^{3} + (- 0.44444 \dots)^{2} - 2 (- 0.44444 \dots) - 1 = - 0.00137 \dots

f^{^{'}} (u_{2}) = 3 (- 0.44444 \dots)^{2} + 2 (- 0.44444 \dots) - 2 = - 2.29629 \dots

u_{3} = - 0.44504 \dots

Δ u_{3} = ∣ - 0.44504 \dots - (- 0.44444 \dots) ∣ = 0.00059 \dots

Δ u_{3} = 0.00059 \dots

u_{4} = - 0.44504 \dots : Δ u_{4} = 5.19031 E - 8

f (u_{3}) = (- 0.44504 \dots)^{3} + (- 0.44504 \dots)^{2} - 2 (- 0.44504 \dots) - 1 = - 1.19164 E - 7

f^{^{'}} (u_{3}) = 3 (- 0.44504 \dots)^{2} + 2 (- 0.44504 \dots) - 2 = - 2.29589 \dots

u_{4} = - 0.44504 \dots

Δ u_{4} = ∣ - 0.44504 \dots - (- 0.44504 \dots) ∣ = 5.19031 E - 8

Δ u_{4} = 5.19031 E - 8

u \approx - 0.44504 \dots

使用长除法

Eq u a t i o n 0 : \frac{u ^{3} + u ^{2} - 2 u - 1}{u + 0.44504 \dots} = u^{2} + 0.55495 \dots u - 2.24697 \dots

u^{2} + 0.55495 \dots u - 2.24697 \dots \approx 0

使用牛顿-拉弗森方法找到

u^{2} + 0.55495 \dots u - 2.24697 \dots = 0

的一个解:

u \approx 1.24697 \dots

u^{2} + 0.55495 \dots u - 2.24697 \dots = 0

牛顿-拉弗森近似法定义

f (u) = u^{2} + 0.55495 \dots u - 2.24697 \dots

找到

f^{^{'}} (u) : 2 u + 0.55495 \dots

\frac{d}{d u} (u^{2} + 0.55495 \dots u - 2.24697 \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} (0.55495 \dots u) - \frac{d}{d u} (2.24697 \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} (0.55495 \dots u) = 0.55495 \dots

\frac{d}{d u} (0.55495 \dots u)

将常数提出:

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

= 0.55495 \dots \frac{d u}{d u}

使用常见微分定则:

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

= 0.55495 \dots \cdot 1

化简

= 0.55495 \dots

\frac{d}{d u} (2.24697 \dots) = 0

\frac{d}{d u} (2.24697 \dots)

常数微分:

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

= 0

= 2 u + 0.55495 \dots - 0

化简

= 2 u + 0.55495 \dots

令

u_{0} = 4

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 2.13291 \dots : Δ u_{1} = 1.86708 \dots

f (u_{0}) = 4^{2} + 0.55495 \dots \cdot 4 - 2.24697 \dots = 15.97285 \dots

f^{^{'}} (u_{0}) = 2 \cdot 4 + 0.55495 \dots = 8.55495 \dots

u_{1} = 2.13291 \dots

Δ u_{1} = ∣ 2.13291 \dots - 4 ∣ = 1.86708 \dots

Δ u_{1} = 1.86708 \dots

u_{2} = 1.40979 \dots : Δ u_{2} = 0.72312 \dots

f (u_{1}) = 2.13291 \dots^{2} + 0.55495 \dots \cdot 2.13291 \dots - 2.24697 \dots = 3.48601 \dots

f^{^{'}} (u_{1}) = 2 \cdot 2.13291 \dots + 0.55495 \dots = 4.82078 \dots

u_{2} = 1.40979 \dots

Δ u_{2} = ∣ 1.40979 \dots - 2.13291 \dots ∣ = 0.72312 \dots

Δ u_{2} = 0.72312 \dots

u_{3} = 1.25483 \dots : Δ u_{3} = 0.15495 \dots

f (u_{2}) = 1.40979 \dots^{2} + 0.55495 \dots \cdot 1.40979 \dots - 2.24697 \dots = 0.52290 \dots

f^{^{'}} (u_{2}) = 2 \cdot 1.40979 \dots + 0.55495 \dots = 3.37453 \dots

u_{3} = 1.25483 \dots

Δ u_{3} = ∣ 1.25483 \dots - 1.40979 \dots ∣ = 0.15495 \dots

Δ u_{3} = 0.15495 \dots

u_{4} = 1.24699 \dots : Δ u_{4} = 0.00783 \dots

f (u_{3}) = 1.25483 \dots^{2} + 0.55495 \dots \cdot 1.25483 \dots - 2.24697 \dots = 0.02401 \dots

f^{^{'}} (u_{3}) = 2 \cdot 1.25483 \dots + 0.55495 \dots = 3.06462 \dots

u_{4} = 1.24699 \dots

Δ u_{4} = ∣ 1.24699 \dots - 1.25483 \dots ∣ = 0.00783 \dots

Δ u_{4} = 0.00783 \dots

u_{5} = 1.24697 \dots : Δ u_{5} = 0.00002 \dots

f (u_{4}) = 1.24699 \dots^{2} + 0.55495 \dots \cdot 1.24699 \dots - 2.24697 \dots = 0.00006 \dots

f^{^{'}} (u_{4}) = 2 \cdot 1.24699 \dots + 0.55495 \dots = 3.04895 \dots

u_{5} = 1.24697 \dots

Δ u_{5} = ∣ 1.24697 \dots - 1.24699 \dots ∣ = 0.00002 \dots

Δ u_{5} = 0.00002 \dots

u_{6} = 1.24697 \dots : Δ u_{6} = 1.32956 E - 10

f (u_{5}) = 1.24697 \dots^{2} + 0.55495 \dots \cdot 1.24697 \dots - 2.24697 \dots = 4.05373 E - 10

f^{^{'}} (u_{5}) = 2 \cdot 1.24697 \dots + 0.55495 \dots = 3.04891 \dots

u_{6} = 1.24697 \dots

Δ u_{6} = ∣ 1.24697 \dots - 1.24697 \dots ∣ = 1.32956 E - 10

Δ u_{6} = 1.32956 E - 10

u \approx 1.24697 \dots

使用长除法

Eq u a t i o n 0 : \frac{u ^{2} + 0.55495 \dots u - 2.24697 \dots}{u - 1.24697 \dots} = u + 1.80193 \dots

u + 1.80193 \dots \approx 0

u \approx - 1.80193 \dots

解为

u \approx - 0.44504 \dots, u \approx 1.24697 \dots, u \approx - 1.80193 \dots

u \approx - 0.44504 \dots, u \approx 1.24697 \dots, u \approx - 1.80193 \dots

验证解

找到无定义的点（奇点）:

u = 0

取

- 2 - \frac{1}{u} + u + 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 - 0.44504 \dots, u \approx 1.24697 \dots, u \approx - 1.80193 \dots

u = sec (x)

代回

sec (x) \approx - 0.44504 \dots, sec (x) \approx 1.24697 \dots, sec (x) \approx - 1.80193 \dots

sec (x) \approx - 0.44504 \dots, sec (x) \approx 1.24697 \dots, sec (x) \approx - 1.80193 \dots

sec (x) = - 0.44504 \dots :

无解

sec (x) = - 0.44504 \dots

sec (x) \leq - 1 or sec (x) \geq 1

无解

sec (x) = 1.24697 \dots : x = arcsec (1.24697 \dots) + 2 πn, x = 2 π - arcsec (1.24697 \dots) + 2 πn

sec (x) = 1.24697 \dots

使用反三角函数性质

sec (x) = 1.24697 \dots

sec (x) = 1.24697 \dots

的通解

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

x = arcsec (1.24697 \dots) + 2 πn, x = 2 π - arcsec (1.24697 \dots) + 2 πn

x = arcsec (1.24697 \dots) + 2 πn, x = 2 π - arcsec (1.24697 \dots) + 2 πn

sec (x) = - 1.80193 \dots : x = arcsec (- 1.80193 \dots) + 2 πn, x = - arcsec (- 1.80193 \dots) + 2 πn

sec (x) = - 1.80193 \dots

使用反三角函数性质

sec (x) = - 1.80193 \dots

sec (x) = - 1.80193 \dots

的通解

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

x = arcsec (- 1.80193 \dots) + 2 πn, x = - arcsec (- 1.80193 \dots) + 2 πn

x = arcsec (- 1.80193 \dots) + 2 πn, x = - arcsec (- 1.80193 \dots) + 2 πn

合并所有解

x = arcsec (1.24697 \dots) + 2 πn, x = 2 π - arcsec (1.24697 \dots) + 2 πn, x = arcsec (- 1.80193 \dots) + 2 πn, x = - arcsec (- 1.80193 \dots) + 2 πn

合并所有解

x = arcsec (1.24697 \dots) + 2 πn, x = 2 π - arcsec (1.24697 \dots) + 2 πn, x = arcsec (- 1.80193 \dots) + 2 πn, x = - arcsec (- 1.80193 \dots) + 2 πn

将解代入原方程进行验证

将它们代入

tan^{2} (x) = \frac{1}{cos ( x ) + 1}

检验解是否符合

去除与方程不符的解。

检验

arcsec (1.24697 \dots) + 2 πn

的解:

真

arcsec (1.24697 \dots) + 2 πn

代入

n = 1

arcsec (1.24697 \dots) + 2 π 1

对于

tan^{2} (x) = \frac{1}{cos ( x ) + 1}

代入

x = arcsec (1.24697 \dots) + 2 π 1

tan^{2} (arcsec (1.24697 \dots) + 2 π 1) = \frac{1}{cos ( arcsec ( 1.24697 \dots ) + 2 π 1 ) + 1}

整理后得

0.55495 \dots = 0.55495 \dots

\Rightarrow 真

检验

2 π - arcsec (1.24697 \dots) + 2 πn

的解:

真

2 π - arcsec (1.24697 \dots) + 2 πn

代入

n = 1

2 π - arcsec (1.24697 \dots) + 2 π 1

对于

tan^{2} (x) = \frac{1}{cos ( x ) + 1}

代入

x = 2 π - arcsec (1.24697 \dots) + 2 π 1

tan^{2} (2 π - arcsec (1.24697 \dots) + 2 π 1) = \frac{1}{cos ( 2 π - arcsec ( 1.24697 \dots ) + 2 π 1 ) + 1}

整理后得

0.55495 \dots = 0.55495 \dots

\Rightarrow 真

检验

arcsec (- 1.80193 \dots) + 2 πn

的解:

真

arcsec (- 1.80193 \dots) + 2 πn

代入

n = 1

arcsec (- 1.80193 \dots) + 2 π 1

对于

tan^{2} (x) = \frac{1}{cos ( x ) + 1}

代入

x = arcsec (- 1.80193 \dots) + 2 π 1

tan^{2} (arcsec (- 1.80193 \dots) + 2 π 1) = \frac{1}{cos ( arcsec ( - 1.80193 \dots ) + 2 π 1 ) + 1}

整理后得

2.24697 \dots = 2.24697 \dots

\Rightarrow 真

检验

- arcsec (- 1.80193 \dots) + 2 πn

的解:

真

- arcsec (- 1.80193 \dots) + 2 πn

代入

n = 1

- arcsec (- 1.80193 \dots) + 2 π 1

对于

tan^{2} (x) = \frac{1}{cos ( x ) + 1}

代入

x = - arcsec (- 1.80193 \dots) + 2 π 1

tan^{2} (- arcsec (- 1.80193 \dots) + 2 π 1) = \frac{1}{cos ( - arcsec ( - 1.80193 \dots ) + 2 π 1 ) + 1}

整理后得

2.24697 \dots = 2.24697 \dots

\Rightarrow 真

x = arcsec (1.24697 \dots) + 2 πn, x = 2 π - arcsec (1.24697 \dots) + 2 πn, x = arcsec (- 1.80193 \dots) + 2 πn, x = - arcsec (- 1.80193 \dots) + 2 πn

以小数形式表示解

x = 0.64026 \dots + 2 πn, x = 2 π - 0.64026 \dots + 2 πn, x = 2.15910 \dots + 2 πn, x = - 2.15910 \dots + 2 πn

作图

流行的例子

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

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

cos^2(x)+cos^4(x)+cos^6(x)=0

cos^{2} (x) + cos^{4} (x) + cos^{6} (x) = 0

sin (x) = \frac{- 1}{4}

sin^4(x)-sin^2(x)=0

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

(cos(t)-4)(2sin^2(t)-1)=0

(cos (t) - 4) (2 sin^{2} (t) - 1) = 0