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受欢迎的三角函数 >

tan^2(x)+5cos(x)-8=0

解答

tan^{2} (x) + 5 cos (x) - 8 = 0

解答

x = 1.88390 \dots + 2 πn, x = - 1.88390 \dots + 2 πn, x = 1.18685 \dots + 2 πn, x = 2 π - 1.18685 \dots + 2 πn

+1

度数

x = 107.93989 \dots^{\circ} + 36 0^{\circ} n, x = - 107.93989 \dots^{\circ} + 36 0^{\circ} n, x = 68.00170 \dots^{\circ} + 36 0^{\circ} n, x = 291.99829 \dots^{\circ} + 36 0^{\circ} n

求解步骤

tan^{2} (x) + 5 cos (x) - 8 = 0

使用三角恒等式改写

- 8 + tan^{2} (x) + 5 cos (x)

使用基本三角恒等式:

tan (x) = \frac{sin ( x )}{cos ( x )}

= - 8 + (\frac{sin ( x )}{cos ( x )})^{2} + 5 cos (x)

使用指数法则:

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

= - 8 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} + 5 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)

= - 8 + \frac{1 - cos ^{2} ( x )}{cos ^{2} ( x )} + 5 cos (x)

- 8 + \frac{1 - cos ^{2} ( x )}{cos ^{2} ( x )} + 5 cos (x) = 0

用替代法求解

- 8 + \frac{1 - cos ^{2} ( x )}{cos ^{2} ( x )} + 5 cos (x) = 0

令

: cos (x) = u

- 8 + \frac{1 - u ^{2}}{u ^{2}} + 5 u = 0

- 8 + \frac{1 - u ^{2}}{u ^{2}} + 5 u = 0 : u \approx - 0.30801 \dots, u \approx 0.37457 \dots, u \approx 1.73344 \dots

- 8 + \frac{1 - u ^{2}}{u ^{2}} + 5 u = 0

在两边乘以

u^{2}

- 8 + \frac{1 - u ^{2}}{u ^{2}} + 5 u = 0

在两边乘以

u^{2}

- 8 u^{2} + \frac{1 - u ^{2}}{u ^{2}} u^{2} + 5 u u^{2} = 0 \cdot u^{2}

化简

- 8 u^{2} + \frac{1 - u ^{2}}{u ^{2}} u^{2} + 5 u u^{2} = 0 \cdot u^{2}

化简

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

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

分式相乘:

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

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

约分:

u^{2}

= 1 - u^{2}

化简

5 u u^{2} : 5 u^{3}

5 u u^{2}

使用指数法则:

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

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

= 5 u^{1 + 2}

数字相加:

1 + 2 = 3

= 5 u^{3}

化简

0 \cdot u^{2} : 0

0 \cdot u^{2}

使用法则

0 \cdot a = 0

= 0

- 8 u^{2} + 1 - u^{2} + 5 u^{3} = 0

化简

- 8 u^{2} + 1 - u^{2} + 5 u^{3} : 5 u^{3} - 9 u^{2} + 1

- 8 u^{2} + 1 - u^{2} + 5 u^{3}

对同类项分组

= 5 u^{3} - 8 u^{2} - u^{2} + 1

同类项相加:

- 8 u^{2} - u^{2} = - 9 u^{2}

= 5 u^{3} - 9 u^{2} + 1

5 u^{3} - 9 u^{2} + 1 = 0

5 u^{3} - 9 u^{2} + 1 = 0

5 u^{3} - 9 u^{2} + 1 = 0

解

5 u^{3} - 9 u^{2} + 1 = 0 : u \approx - 0.30801 \dots, u \approx 0.37457 \dots, u \approx 1.73344 \dots

5 u^{3} - 9 u^{2} + 1 = 0

使用牛顿-拉弗森方法找到

5 u^{3} - 9 u^{2} + 1 = 0

的一个解:

u \approx - 0.30801 \dots

5 u^{3} - 9 u^{2} + 1 = 0

牛顿-拉弗森近似法定义

f (u) = 5 u^{3} - 9 u^{2} + 1

找到

f^{^{'}} (u) : 15 u^{2} - 18 u

\frac{d}{d u} (5 u^{3} - 9 u^{2} + 1)

使用微分加减法定则:

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

= \frac{d}{d u} (5 u^{3}) - \frac{d}{d u} (9 u^{2}) + \frac{d}{d u} (1)

\frac{d}{d u} (5 u^{3}) = 15 u^{2}

\frac{d}{d u} (5 u^{3})

将常数提出:

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

= 5 \frac{d}{d u} (u^{3})

使用幂法则:

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

= 5 \cdot 3 u^{3 - 1}

化简

= 15 u^{2}

\frac{d}{d u} (9 u^{2}) = 18 u

\frac{d}{d u} (9 u^{2})

将常数提出:

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

= 9 \frac{d}{d u} (u^{2})

使用幂法则:

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

= 9 \cdot 2 u^{2 - 1}

化简

= 18 u

\frac{d}{d u} (1) = 0

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

常数微分:

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

= 0

= 15 u^{2} - 18 u + 0

化简

= 15 u^{2} - 18 u

令

u_{0} = - 1

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = - 0.60606 \dots : Δ u_{1} = 0.39393 \dots

f (u_{0}) = 5 (- 1)^{3} - 9 (- 1)^{2} + 1 = - 13

f^{^{'}} (u_{0}) = 15 (- 1)^{2} - 18 (- 1) = 33

u_{1} = - 0.60606 \dots

Δ u_{1} = ∣ - 0.60606 \dots - (- 1) ∣ = 0.39393 \dots

Δ u_{1} = 0.39393 \dots

u_{2} = - 0.39783 \dots : Δ u_{2} = 0.20822 \dots

f (u_{1}) = 5 (- 0.60606 \dots)^{3} - 9 (- 0.60606 \dots)^{2} + 1 = - 3.41884 \dots

f^{^{'}} (u_{1}) = 15 (- 0.60606 \dots)^{2} - 18 (- 0.60606 \dots) = 16.41873 \dots

u_{2} = - 0.39783 \dots

Δ u_{2} = ∣ - 0.39783 \dots - (- 0.60606 \dots) ∣ = 0.20822 \dots

Δ u_{2} = 0.20822 \dots

u_{3} = - 0.32030 \dots : Δ u_{3} = 0.07753 \dots

f (u_{2}) = 5 (- 0.39783 \dots)^{3} - 9 (- 0.39783 \dots)^{2} + 1 = - 0.73926 \dots

f^{^{'}} (u_{2}) = 15 (- 0.39783 \dots)^{2} - 18 (- 0.39783 \dots) = 9.53504 \dots

u_{3} = - 0.32030 \dots

Δ u_{3} = ∣ - 0.32030 \dots - (- 0.39783 \dots) ∣ = 0.07753 \dots

Δ u_{3} = 0.07753 \dots

u_{4} = - 0.30830 \dots : Δ u_{4} = 0.01199 \dots

f (u_{3}) = 5 (- 0.32030 \dots)^{3} - 9 (- 0.32030 \dots)^{2} + 1 = - 0.08764 \dots

f^{^{'}} (u_{3}) = 15 (- 0.32030 \dots)^{2} - 18 (- 0.32030 \dots) = 7.30431 \dots

u_{4} = - 0.30830 \dots

Δ u_{4} = ∣ - 0.30830 \dots - (- 0.32030 \dots) ∣ = 0.01199 \dots

Δ u_{4} = 0.01199 \dots

u_{5} = - 0.30801 \dots : Δ u_{5} = 0.00028 \dots

f (u_{4}) = 5 (- 0.30830 \dots)^{3} - 9 (- 0.30830 \dots)^{2} + 1 = - 0.00197 \dots

f^{^{'}} (u_{4}) = 15 (- 0.30830 \dots)^{2} - 18 (- 0.30830 \dots) = 6.97521 \dots

u_{5} = - 0.30801 \dots

Δ u_{5} = ∣ - 0.30801 \dots - (- 0.30830 \dots) ∣ = 0.00028 \dots

Δ u_{5} = 0.00028 \dots

u_{6} = - 0.30801 \dots : Δ u_{6} = 1.5734 E - 7

f (u_{5}) = 5 (- 0.30801 \dots)^{3} - 9 (- 0.30801 \dots)^{2} + 1 = - 1.09626 E - 6

f^{^{'}} (u_{5}) = 15 (- 0.30801 \dots)^{2} - 18 (- 0.30801 \dots) = 6.96748 \dots

u_{6} = - 0.30801 \dots

Δ u_{6} = ∣ - 0.30801 \dots - (- 0.30801 \dots) ∣ = 1.5734 E - 7

Δ u_{6} = 1.5734 E - 7

u \approx - 0.30801 \dots

使用长除法

Eq u a t i o n 0 : \frac{5 u ^{3} - 9 u ^{2} + 1}{u + 0.30801 \dots} = 5 u^{2} - 10.54009 \dots u + 3.24655 \dots

5 u^{2} - 10.54009 \dots u + 3.24655 \dots \approx 0

使用牛顿-拉弗森方法找到

5 u^{2} - 10.54009 \dots u + 3.24655 \dots = 0

的一个解:

u \approx 0.37457 \dots

5 u^{2} - 10.54009 \dots u + 3.24655 \dots = 0

牛顿-拉弗森近似法定义

f (u) = 5 u^{2} - 10.54009 \dots u + 3.24655 \dots

找到

f^{^{'}} (u) : 10 u - 10.54009 \dots

\frac{d}{d u} (5 u^{2} - 10.54009 \dots u + 3.24655 \dots)

使用微分加减法定则:

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

= \frac{d}{d u} (5 u^{2}) - \frac{d}{d u} (10.54009 \dots u) + \frac{d}{d u} (3.24655 \dots)

\frac{d}{d u} (5 u^{2}) = 10 u

\frac{d}{d u} (5 u^{2})

将常数提出:

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

= 5 \frac{d}{d u} (u^{2})

使用幂法则:

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

= 5 \cdot 2 u^{2 - 1}

化简

= 10 u

\frac{d}{d u} (10.54009 \dots u) = 10.54009 \dots

\frac{d}{d u} (10.54009 \dots u)

将常数提出:

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

= 10.54009 \dots \frac{d u}{d u}

使用常见微分定则:

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

= 10.54009 \dots \cdot 1

化简

= 10.54009 \dots

\frac{d}{d u} (3.24655 \dots) = 0

\frac{d}{d u} (3.24655 \dots)

常数微分:

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

= 0

= 10 u - 10.54009 \dots + 0

化简

= 10 u - 10.54009 \dots

令

u_{0} = 0

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 0.30801 \dots : Δ u_{1} = 0.30801 \dots

f (u_{0}) = 5 \cdot 0^{2} - 10.54009 \dots \cdot 0 + 3.24655 \dots = 3.24655 \dots

f^{^{'}} (u_{0}) = 10 \cdot 0 - 10.54009 \dots = - 10.54009 \dots

u_{1} = 0.30801 \dots

Δ u_{1} = ∣ 0.30801 \dots - 0 ∣ = 0.30801 \dots

Δ u_{1} = 0.30801 \dots

u_{2} = 0.37160 \dots : Δ u_{2} = 0.06359 \dots

f (u_{1}) = 5 \cdot 0.30801 \dots^{2} - 10.54009 \dots \cdot 0.30801 \dots + 3.24655 \dots = 0.47437 \dots

f^{^{'}} (u_{1}) = 10 \cdot 0.30801 \dots - 10.54009 \dots = - 7.45990 \dots

u_{2} = 0.37160 \dots

Δ u_{2} = ∣ 0.37160 \dots - 0.30801 \dots ∣ = 0.06359 \dots

Δ u_{2} = 0.06359 \dots

u_{3} = 0.37457 \dots : Δ u_{3} = 0.00296 \dots

f (u_{2}) = 5 \cdot 0.37160 \dots^{2} - 10.54009 \dots \cdot 0.37160 \dots + 3.24655 \dots = 0.02021 \dots

f^{^{'}} (u_{2}) = 10 \cdot 0.37160 \dots - 10.54009 \dots = - 6.82399 \dots

u_{3} = 0.37457 \dots

Δ u_{3} = ∣ 0.37457 \dots - 0.37160 \dots ∣ = 0.00296 \dots

Δ u_{3} = 0.00296 \dots

u_{4} = 0.37457 \dots : Δ u_{4} = 6.46028 E - 6

f (u_{3}) = 5 \cdot 0.37457 \dots^{2} - 10.54009 \dots \cdot 0.37457 \dots + 3.24655 \dots = 0.00004 \dots

f^{^{'}} (u_{3}) = 10 \cdot 0.37457 \dots - 10.54009 \dots = - 6.79437 \dots

u_{4} = 0.37457 \dots

Δ u_{4} = ∣ 0.37457 \dots - 0.37457 \dots ∣ = 6.46028 E - 6

Δ u_{4} = 6.46028 E - 6

u_{5} = 0.37457 \dots : Δ u_{5} = 3.07134 E - 11

f (u_{4}) = 5 \cdot 0.37457 \dots^{2} - 10.54009 \dots \cdot 0.37457 \dots + 3.24655 \dots = 2.08676 E - 10

f^{^{'}} (u_{4}) = 10 \cdot 0.37457 \dots - 10.54009 \dots = - 6.79430 \dots

u_{5} = 0.37457 \dots

Δ u_{5} = ∣ 0.37457 \dots - 0.37457 \dots ∣ = 3.07134 E - 11

Δ u_{5} = 3.07134 E - 11

u \approx 0.37457 \dots

使用长除法

Eq u a t i o n 0 : \frac{5 u ^{2} - 10.54009 \dots u + 3.24655 \dots}{u - 0.37457 \dots} = 5 u - 8.66720 \dots

5 u - 8.66720 \dots \approx 0

u \approx 1.73344 \dots

解为

u \approx - 0.30801 \dots, u \approx 0.37457 \dots, u \approx 1.73344 \dots

u \approx - 0.30801 \dots, u \approx 0.37457 \dots, u \approx 1.73344 \dots

验证解

找到无定义的点（奇点）:

u = 0

取

- 8 + \frac{1 - u ^{2}}{u ^{2}} + 5 u

的分母，令其等于零

解

u^{2} = 0 : u = 0

u^{2} = 0

使用法则

x^{n} = 0 \Rightarrow x = 0

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 - 0.30801 \dots, u \approx 0.37457 \dots, u \approx 1.73344 \dots

u = cos (x)

代回

cos (x) \approx - 0.30801 \dots, cos (x) \approx 0.37457 \dots, cos (x) \approx 1.73344 \dots

cos (x) \approx - 0.30801 \dots, cos (x) \approx 0.37457 \dots, cos (x) \approx 1.73344 \dots

cos (x) = - 0.30801 \dots : x = arccos (- 0.30801 \dots) + 2 πn, x = - arccos (- 0.30801 \dots) + 2 πn

cos (x) = - 0.30801 \dots

使用反三角函数性质

cos (x) = - 0.30801 \dots

cos (x) = - 0.30801 \dots

的通解

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

x = arccos (- 0.30801 \dots) + 2 πn, x = - arccos (- 0.30801 \dots) + 2 πn

x = arccos (- 0.30801 \dots) + 2 πn, x = - arccos (- 0.30801 \dots) + 2 πn

cos (x) = 0.37457 \dots : x = arccos (0.37457 \dots) + 2 πn, x = 2 π - arccos (0.37457 \dots) + 2 πn

cos (x) = 0.37457 \dots

使用反三角函数性质

cos (x) = 0.37457 \dots

cos (x) = 0.37457 \dots

的通解

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

x = arccos (0.37457 \dots) + 2 πn, x = 2 π - arccos (0.37457 \dots) + 2 πn

x = arccos (0.37457 \dots) + 2 πn, x = 2 π - arccos (0.37457 \dots) + 2 πn

cos (x) = 1.73344 \dots :

无解

cos (x) = 1.73344 \dots

- 1 \leq cos (x) \leq 1

无解

合并所有解

x = arccos (- 0.30801 \dots) + 2 πn, x = - arccos (- 0.30801 \dots) + 2 πn, x = arccos (0.37457 \dots) + 2 πn, x = 2 π - arccos (0.37457 \dots) + 2 πn

以小数形式表示解

x = 1.88390 \dots + 2 πn, x = - 1.88390 \dots + 2 πn, x = 1.18685 \dots + 2 πn, x = 2 π - 1.18685 \dots + 2 πn

作图

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