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受欢迎的三角函数 >

tan(x)=2+tan^3(x)

解答

tan (x) = 2 + tan^{3} (x)

解答

x = - 0.98930 \dots + πn

+1

度数

x = - 56.68315 \dots^{\circ} + 18 0^{\circ} n

求解步骤

tan (x) = 2 + tan^{3} (x)

用替代法求解

tan (x) = 2 + tan^{3} (x)

令

: tan (x) = u

u = 2 + u^{3}

u = 2 + u^{3} : u \approx - 1.52137 \dots

u = 2 + u^{3}

交换两边

2 + u^{3} = u

将

u

para o lado esquerdo

2 + u^{3} = u

两边减去

u

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

化简

2 + u^{3} - u = 0

2 + u^{3} - u = 0

改写成标准形式

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

u^{3} - u + 2 = 0

使用牛顿-拉弗森方法找到

u^{3} - u + 2 = 0

的一个解:

u \approx - 1.52137 \dots

u^{3} - u + 2 = 0

牛顿-拉弗森近似法定义

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

找到

f^{^{'}} (u) : 3 u^{2} - 1

\frac{d}{d u} (u^{3} - 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 u}{d 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 u}{d u} = 1

\frac{d u}{d u}

使用常见微分定则:

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

= 1

\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} - 1 + 0

化简

= 3 u^{2} - 1

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

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

u_{1} = - 2

Δ u_{1} = ∣ - 2 - (- 1) ∣ = 1

Δ u_{1} = 1

u_{2} = - 1.63636 \dots : Δ u_{2} = 0.36363 \dots

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

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

u_{2} = - 1.63636 \dots

Δ u_{2} = ∣ - 1.63636 \dots - (- 2) ∣ = 0.36363 \dots

Δ u_{2} = 0.36363 \dots

u_{3} = - 1.53039 \dots : Δ u_{3} = 0.10597 \dots

f (u_{2}) = (- 1.63636 \dots)^{3} - (- 1.63636 \dots) + 2 = - 0.74530 \dots

f^{^{'}} (u_{2}) = 3 (- 1.63636 \dots)^{2} - 1 = 7.03305 \dots

u_{3} = - 1.53039 \dots

Δ u_{3} = ∣ - 1.53039 \dots - (- 1.63636 \dots) ∣ = 0.10597 \dots

Δ u_{3} = 0.10597 \dots

u_{4} = - 1.52144 \dots : Δ u_{4} = 0.00895 \dots

f (u_{3}) = (- 1.53039 \dots)^{3} - (- 1.53039 \dots) + 2 = - 0.05393 \dots

f^{^{'}} (u_{3}) = 3 (- 1.53039 \dots)^{2} - 1 = 6.02629 \dots

u_{4} = - 1.52144 \dots

Δ u_{4} = ∣ - 1.52144 \dots - (- 1.53039 \dots) ∣ = 0.00895 \dots

Δ u_{4} = 0.00895 \dots

u_{5} = - 1.52137 \dots : Δ u_{5} = 0.00006 \dots

f (u_{4}) = (- 1.52144 \dots)^{3} - (- 1.52144 \dots) + 2 = - 0.00036 \dots

f^{^{'}} (u_{4}) = 3 (- 1.52144 \dots)^{2} - 1 = 5.94435 \dots

u_{5} = - 1.52137 \dots

Δ u_{5} = ∣ - 1.52137 \dots - (- 1.52144 \dots) ∣ = 0.00006 \dots

Δ u_{5} = 0.00006 \dots

u_{6} = - 1.52137 \dots : Δ u_{6} = 2.92858 E - 9

f (u_{5}) = (- 1.52137 \dots)^{3} - (- 1.52137 \dots) + 2 = - 1.74069 E - 8

f^{^{'}} (u_{5}) = 3 (- 1.52137 \dots)^{2} - 1 = 5.94378 \dots

u_{6} = - 1.52137 \dots

Δ u_{6} = ∣ - 1.52137 \dots - (- 1.52137 \dots) ∣ = 2.92858 E - 9

Δ u_{6} = 2.92858 E - 9

u \approx - 1.52137 \dots

使用长除法

Eq u a t i o n 0 : \frac{u ^{3} - u + 2}{u + 1.52137 \dots} = u^{2} - 1.52137 \dots u + 1.31459 \dots

u^{2} - 1.52137 \dots u + 1.31459 \dots \approx 0

使用牛顿-拉弗森方法找到

u^{2} - 1.52137 \dots u + 1.31459 \dots = 0

的一个解:

u \in R

无解

u^{2} - 1.52137 \dots u + 1.31459 \dots = 0

牛顿-拉弗森近似法定义

f (u) = u^{2} - 1.52137 \dots u + 1.31459 \dots

找到

f^{^{'}} (u) : 2 u - 1.52137 \dots

\frac{d}{d u} (u^{2} - 1.52137 \dots u + 1.31459 \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} (1.52137 \dots u) + \frac{d}{d u} (1.31459 \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} (1.52137 \dots u) = 1.52137 \dots

\frac{d}{d u} (1.52137 \dots u)

将常数提出:

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

= 1.52137 \dots \frac{d u}{d u}

使用常见微分定则:

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

= 1.52137 \dots \cdot 1

化简

= 1.52137 \dots

\frac{d}{d u} (1.31459 \dots) = 0

\frac{d}{d u} (1.31459 \dots)

常数微分:

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

= 0

= 2 u - 1.52137 \dots + 0

化简

= 2 u - 1.52137 \dots

令

u_{0} = 1

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = - 0.65729 \dots : Δ u_{1} = 1.65729 \dots

f (u_{0}) = 1^{2} - 1.52137 \dots \cdot 1 + 1.31459 \dots = 0.79321 \dots

f^{^{'}} (u_{0}) = 2 \cdot 1 - 1.52137 \dots = 0.47862 \dots

u_{1} = - 0.65729 \dots

Δ u_{1} = ∣ - 0.65729 \dots - 1 ∣ = 1.65729 \dots

Δ u_{1} = 1.65729 \dots

u_{2} = 0.31119 \dots : Δ u_{2} = 0.96849 \dots

f (u_{1}) = (- 0.65729 \dots)^{2} - 1.52137 \dots (- 0.65729 \dots) + 1.31459 \dots = 2.74663 \dots

f^{^{'}} (u_{1}) = 2 (- 0.65729 \dots) - 1.52137 \dots = - 2.83597 \dots

u_{2} = 0.31119 \dots

Δ u_{2} = ∣ 0.31119 \dots - (- 0.65729 \dots) ∣ = 0.96849 \dots

Δ u_{2} = 0.96849 \dots

u_{3} = 1.35459 \dots : Δ u_{3} = 1.04339 \dots

f (u_{2}) = 0.31119 \dots^{2} - 1.52137 \dots \cdot 0.31119 \dots + 1.31459 \dots = 0.93798 \dots

f^{^{'}} (u_{2}) = 2 \cdot 0.31119 \dots - 1.52137 \dots = - 0.89897 \dots

u_{3} = 1.35459 \dots

Δ u_{3} = ∣ 1.35459 \dots - 0.31119 \dots ∣ = 1.04339 \dots

Δ u_{3} = 1.04339 \dots

u_{4} = 0.43805 \dots : Δ u_{4} = 0.91653 \dots

f (u_{3}) = 1.35459 \dots^{2} - 1.52137 \dots \cdot 1.35459 \dots + 1.31459 \dots = 1.08866 \dots

f^{^{'}} (u_{3}) = 2 \cdot 1.35459 \dots - 1.52137 \dots = 1.18780 \dots

u_{4} = 0.43805 \dots

Δ u_{4} = ∣ 0.43805 \dots - 1.35459 \dots ∣ = 0.91653 \dots

Δ u_{4} = 0.91653 \dots

u_{5} = 1.73989 \dots : Δ u_{5} = 1.30184 \dots

f (u_{4}) = 0.43805 \dots^{2} - 1.52137 \dots \cdot 0.43805 \dots + 1.31459 \dots = 0.84004 \dots

f^{^{'}} (u_{4}) = 2 \cdot 0.43805 \dots - 1.52137 \dots = - 0.64527 \dots

u_{5} = 1.73989 \dots

Δ u_{5} = ∣ 1.73989 \dots - 0.43805 \dots ∣ = 1.30184 \dots

Δ u_{5} = 1.30184 \dots

u_{6} = 0.87450 \dots : Δ u_{6} = 0.86539 \dots

f (u_{5}) = 1.73989 \dots^{2} - 1.52137 \dots \cdot 1.73989 \dots + 1.31459 \dots = 1.69479 \dots

f^{^{'}} (u_{5}) = 2 \cdot 1.73989 \dots - 1.52137 \dots = 1.95841 \dots

u_{6} = 0.87450 \dots

Δ u_{6} = ∣ 0.87450 \dots - 1.73989 \dots ∣ = 0.86539 \dots

Δ u_{6} = 0.86539 \dots

u_{7} = - 2.41546 \dots : Δ u_{7} = 3.28996 \dots

f (u_{6}) = 0.87450 \dots^{2} - 1.52137 \dots \cdot 0.87450 \dots + 1.31459 \dots = 0.74890 \dots

f^{^{'}} (u_{6}) = 2 \cdot 0.87450 \dots - 1.52137 \dots = 0.22763 \dots

u_{7} = - 2.41546 \dots

Δ u_{7} = ∣ - 2.41546 \dots - 0.87450 \dots ∣ = 3.28996 \dots

Δ u_{7} = 3.28996 \dots

u_{8} = - 0.71153 \dots : Δ u_{8} = 1.70393 \dots

f (u_{7}) = (- 2.41546 \dots)^{2} - 1.52137 \dots (- 2.41546 \dots) + 1.31459 \dots = 10.82387 \dots

f^{^{'}} (u_{7}) = 2 (- 2.41546 \dots) - 1.52137 \dots = - 6.35230 \dots

u_{8} = - 0.71153 \dots

Δ u_{8} = ∣ - 0.71153 \dots - (- 2.41546 \dots) ∣ = 1.70393 \dots

Δ u_{8} = 1.70393 \dots

u_{9} = 0.27452 \dots : Δ u_{9} = 0.98605 \dots

f (u_{8}) = (- 0.71153 \dots)^{2} - 1.52137 \dots (- 0.71153 \dots) + 1.31459 \dots = 2.90337 \dots

f^{^{'}} (u_{8}) = 2 (- 0.71153 \dots) - 1.52137 \dots = - 2.94443 \dots

u_{9} = 0.27452 \dots

Δ u_{9} = ∣ 0.27452 \dots - (- 0.71153 \dots) ∣ = 0.98605 \dots

Δ u_{9} = 0.98605 \dots

u_{10} = 1.27449 \dots : Δ u_{10} = 0.99997 \dots

f (u_{9}) = 0.27452 \dots^{2} - 1.52137 \dots \cdot 0.27452 \dots + 1.31459 \dots = 0.97230 \dots

f^{^{'}} (u_{9}) = 2 \cdot 0.27452 \dots - 1.52137 \dots = - 0.97233 \dots

u_{10} = 1.27449 \dots

Δ u_{10} = ∣ 1.27449 \dots - 0.27452 \dots ∣ = 0.99997 \dots

Δ u_{10} = 0.99997 \dots

无法得出解

解是

u \approx - 1.52137 \dots

u = tan (x)

代回

tan (x) \approx - 1.52137 \dots

tan (x) \approx - 1.52137 \dots

tan (x) = - 1.52137 \dots : x = arctan (- 1.52137 \dots) + πn

tan (x) = - 1.52137 \dots

使用反三角函数性质

tan (x) = - 1.52137 \dots

tan (x) = - 1.52137 \dots

的通解

tan (x) = - a \Rightarrow x = arctan (- a) + πn

x = arctan (- 1.52137 \dots) + πn

x = arctan (- 1.52137 \dots) + πn

合并所有解

x = arctan (- 1.52137 \dots) + πn

以小数形式表示解

x = - 0.98930 \dots + πn

作图

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