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4cos^3(x)-3cos(x)=-0.7071

解答

4 cos^{3} (x) - 3 cos (x) = - 0.7071

解答

x = 1.30900 \dots + 2 πn, x = 2 π - 1.30900 \dots + 2 πn, x = 0.78539 \dots + 2 πn, x = 2 π - 0.78539 \dots + 2 πn, x = 2.87979 \dots + 2 πn, x = - 2.87979 \dots + 2 πn

+1

度数

x = 75.00018 \dots^{\circ} + 36 0^{\circ} n, x = 284.99981 \dots^{\circ} + 36 0^{\circ} n, x = 44.99981 \dots^{\circ} + 36 0^{\circ} n, x = 315.00018 \dots^{\circ} + 36 0^{\circ} n, x = 164.99981 \dots^{\circ} + 36 0^{\circ} n, x = - 164.99981 \dots^{\circ} + 36 0^{\circ} n

求解步骤

4 cos^{3} (x) - 3 cos (x) = - 0.7071

用替代法求解

4 cos^{3} (x) - 3 cos (x) = - 0.7071

令

: cos (x) = u

4 u^{3} - 3 u = - 0.7071

4 u^{3} - 3 u = - 0.7071 : u \approx 0.25881 \dots, u \approx 0.70710 \dots, u \approx - 0.96592 \dots

4 u^{3} - 3 u = - 0.7071

将

0.7071

para o lado esquerdo

4 u^{3} - 3 u = - 0.7071

两边加上

0.7071

4 u^{3} - 3 u + 0.7071 = - 0.7071 + 0.7071

化简

4 u^{3} - 3 u + 0.7071 = 0

4 u^{3} - 3 u + 0.7071 = 0

使用牛顿-拉弗森方法找到

4 u^{3} - 3 u + 0.7071 = 0

的一个解:

u \approx 0.25881 \dots

4 u^{3} - 3 u + 0.7071 = 0

牛顿-拉弗森近似法定义

f (u) = 4 u^{3} - 3 u + 0.7071

找到

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

\frac{d}{d u} (4 u^{3} - 3 u + 0.7071)

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} (3 u) + \frac{d}{d u} (0.7071)

\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} (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} (0.7071) = 0

\frac{d}{d u} (0.7071)

常数微分:

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

= 0

= 12 u^{2} - 3 + 0

化简

= 12 u^{2} - 3

令

u_{0} = 0

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 0.2357 : Δ u_{1} = 0.2357

f (u_{0}) = 4 \cdot 0^{3} - 3 \cdot 0 + 0.7071 = 0.7071

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

u_{1} = 0.2357

Δ u_{1} = ∣ 0.2357 - 0 ∣ = 0.2357

Δ u_{1} = 0.2357

u_{2} = 0.25814 \dots : Δ u_{2} = 0.02244 \dots

f (u_{1}) = 4 \cdot 0.235 7^{3} - 3 \cdot 0.2357 + 0.7071 = 0.05237 \dots

f^{^{'}} (u_{1}) = 12 \cdot 0.235 7^{2} - 3 = - 2.33334612

u_{2} = 0.25814 \dots

Δ u_{2} = ∣ 0.25814 \dots - 0.2357 ∣ = 0.02244 \dots

Δ u_{2} = 0.02244 \dots

u_{3} = 0.25881 \dots : Δ u_{3} = 0.00066 \dots

f (u_{2}) = 4 \cdot 0.25814 \dots^{3} - 3 \cdot 0.25814 \dots + 0.7071 = 0.00147 \dots

f^{^{'}} (u_{2}) = 12 \cdot 0.25814 \dots^{2} - 3 = - 2.20032 \dots

u_{3} = 0.25881 \dots

Δ u_{3} = ∣ 0.25881 \dots - 0.25814 \dots ∣ = 0.00066 \dots

Δ u_{3} = 0.00066 \dots

u_{4} = 0.25881 \dots : Δ u_{4} = 6.30447 E - 7

f (u_{3}) = 4 \cdot 0.25881 \dots^{3} - 3 \cdot 0.25881 \dots + 0.7071 = 1.38457 E - 6

f^{^{'}} (u_{3}) = 12 \cdot 0.25881 \dots^{2} - 3 = - 2.19617 \dots

u_{4} = 0.25881 \dots

Δ u_{4} = ∣ 0.25881 \dots - 0.25881 \dots ∣ = 6.30447 E - 7

Δ u_{4} = 6.30447 E - 7

u \approx 0.25881 \dots

使用长除法

Eq u a t i o n 0 : \frac{4 u ^{3} - 3 u + 0.7071}{u - 0.25881 \dots} = 4 u^{2} + 1.03526 \dots u - 2.73205 \dots

4 u^{2} + 1.03526 \dots u - 2.73205 \dots \approx 0

使用牛顿-拉弗森方法找到

4 u^{2} + 1.03526 \dots u - 2.73205 \dots = 0

的一个解:

u \approx 0.70710 \dots

4 u^{2} + 1.03526 \dots u - 2.73205 \dots = 0

牛顿-拉弗森近似法定义

f (u) = 4 u^{2} + 1.03526 \dots u - 2.73205 \dots

找到

f^{^{'}} (u) : 8 u + 1.03526 \dots

\frac{d}{d u} (4 u^{2} + 1.03526 \dots u - 2.73205 \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} (1.03526 \dots u) - \frac{d}{d u} (2.73205 \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} (1.03526 \dots u) = 1.03526 \dots

\frac{d}{d u} (1.03526 \dots u)

将常数提出:

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

= 1.03526 \dots \frac{d u}{d u}

使用常见微分定则:

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

= 1.03526 \dots \cdot 1

化简

= 1.03526 \dots

\frac{d}{d u} (2.73205 \dots) = 0

\frac{d}{d u} (2.73205 \dots)

常数微分:

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

= 0

= 8 u + 1.03526 \dots - 0

化简

= 8 u + 1.03526 \dots

令

u_{0} = 3

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 1.54710 \dots : Δ u_{1} = 1.45289 \dots

f (u_{0}) = 4 \cdot 3^{2} + 1.03526 \dots \cdot 3 - 2.73205 \dots = 36.37373 \dots

f^{^{'}} (u_{0}) = 8 \cdot 3 + 1.03526 \dots = 25.03526 \dots

u_{1} = 1.54710 \dots

Δ u_{1} = ∣ 1.54710 \dots - 3 ∣ = 1.45289 \dots

Δ u_{1} = 1.45289 \dots

u_{2} = 0.91754 \dots : Δ u_{2} = 0.62955 \dots

f (u_{1}) = 4 \cdot 1.54710 \dots^{2} + 1.03526 \dots \cdot 1.54710 \dots - 2.73205 \dots = 8.44367 \dots

f^{^{'}} (u_{1}) = 8 \cdot 1.54710 \dots + 1.03526 \dots = 13.41206 \dots

u_{2} = 0.91754 \dots

Δ u_{2} = ∣ 0.91754 \dots - 1.54710 \dots ∣ = 0.62955 \dots

Δ u_{2} = 0.62955 \dots

u_{3} = 0.72825 \dots : Δ u_{3} = 0.18928 \dots

f (u_{2}) = 4 \cdot 0.91754 \dots^{2} + 1.03526 \dots \cdot 0.91754 \dots - 2.73205 \dots = 1.58537 \dots

f^{^{'}} (u_{2}) = 8 \cdot 0.91754 \dots + 1.03526 \dots = 8.37559 \dots

u_{3} = 0.72825 \dots

Δ u_{3} = ∣ 0.72825 \dots - 0.91754 \dots ∣ = 0.18928 \dots

Δ u_{3} = 0.18928 \dots

u_{4} = 0.70736 \dots : Δ u_{4} = 0.02088 \dots

f (u_{3}) = 4 \cdot 0.72825 \dots^{2} + 1.03526 \dots \cdot 0.72825 \dots - 2.73205 \dots = 0.14331 \dots

f^{^{'}} (u_{3}) = 8 \cdot 0.72825 \dots + 1.03526 \dots = 6.86132 \dots

u_{4} = 0.70736 \dots

Δ u_{4} = ∣ 0.70736 \dots - 0.72825 \dots ∣ = 0.02088 \dots

Δ u_{4} = 0.02088 \dots

u_{5} = 0.70710 \dots : Δ u_{5} = 0.00026 \dots

f (u_{4}) = 4 \cdot 0.70736 \dots^{2} + 1.03526 \dots \cdot 0.70736 \dots - 2.73205 \dots = 0.00174 \dots

f^{^{'}} (u_{4}) = 8 \cdot 0.70736 \dots + 1.03526 \dots = 6.69422 \dots

u_{5} = 0.70710 \dots

Δ u_{5} = ∣ 0.70710 \dots - 0.70736 \dots ∣ = 0.00026 \dots

Δ u_{5} = 0.00026 \dots

u_{6} = 0.70710 \dots : Δ u_{6} = 4.06209 E - 8

f (u_{5}) = 4 \cdot 0.70710 \dots^{2} + 1.03526 \dots \cdot 0.70710 \dots - 2.73205 \dots = 2.71841 E - 7

f^{^{'}} (u_{5}) = 8 \cdot 0.70710 \dots + 1.03526 \dots = 6.69213 \dots

u_{6} = 0.70710 \dots

Δ u_{6} = ∣ 0.70710 \dots - 0.70710 \dots ∣ = 4.06209 E - 8

Δ u_{6} = 4.06209 E - 8

u \approx 0.70710 \dots

使用长除法

Eq u a t i o n 0 : \frac{4 u ^{2} + 1.03526 \dots u - 2.73205 \dots}{u - 0.70710 \dots} = 4 u + 3.86369 \dots

4 u + 3.86369 \dots \approx 0

u \approx - 0.96592 \dots

解为

u \approx 0.25881 \dots, u \approx 0.70710 \dots, u \approx - 0.96592 \dots

u = cos (x)

代回

cos (x) \approx 0.25881 \dots, cos (x) \approx 0.70710 \dots, cos (x) \approx - 0.96592 \dots

cos (x) \approx 0.25881 \dots, cos (x) \approx 0.70710 \dots, cos (x) \approx - 0.96592 \dots

cos (x) = 0.25881 \dots : x = arccos (0.25881 \dots) + 2 πn, x = 2 π - arccos (0.25881 \dots) + 2 πn

cos (x) = 0.25881 \dots

使用反三角函数性质

cos (x) = 0.25881 \dots

cos (x) = 0.25881 \dots

的通解

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

x = arccos (0.25881 \dots) + 2 πn, x = 2 π - arccos (0.25881 \dots) + 2 πn

x = arccos (0.25881 \dots) + 2 πn, x = 2 π - arccos (0.25881 \dots) + 2 πn

cos (x) = 0.70710 \dots : x = arccos (0.70710 \dots) + 2 πn, x = 2 π - arccos (0.70710 \dots) + 2 πn

cos (x) = 0.70710 \dots

使用反三角函数性质

cos (x) = 0.70710 \dots

cos (x) = 0.70710 \dots

的通解

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

x = arccos (0.70710 \dots) + 2 πn, x = 2 π - arccos (0.70710 \dots) + 2 πn

x = arccos (0.70710 \dots) + 2 πn, x = 2 π - arccos (0.70710 \dots) + 2 πn

cos (x) = - 0.96592 \dots : x = arccos (- 0.96592 \dots) + 2 πn, x = - arccos (- 0.96592 \dots) + 2 πn

cos (x) = - 0.96592 \dots

使用反三角函数性质

cos (x) = - 0.96592 \dots

cos (x) = - 0.96592 \dots

的通解

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

x = arccos (- 0.96592 \dots) + 2 πn, x = - arccos (- 0.96592 \dots) + 2 πn

x = arccos (- 0.96592 \dots) + 2 πn, x = - arccos (- 0.96592 \dots) + 2 πn

合并所有解

x = arccos (0.25881 \dots) + 2 πn, x = 2 π - arccos (0.25881 \dots) + 2 πn, x = arccos (0.70710 \dots) + 2 πn, x = 2 π - arccos (0.70710 \dots) + 2 πn, x = arccos (- 0.96592 \dots) + 2 πn, x = - arccos (- 0.96592 \dots) + 2 πn

以小数形式表示解

x = 1.30900 \dots + 2 πn, x = 2 π - 1.30900 \dots + 2 πn, x = 0.78539 \dots + 2 πn, x = 2 π - 0.78539 \dots + 2 πn, x = 2.87979 \dots + 2 πn, x = - 2.87979 \dots + 2 πn

流行的例子

2tan(x)=tan(2x)

2 tan (x) = tan (2 x)

sin^2(θ)=3(cos(-θ)+1)

sin^{2} (θ) = 3 (cos (- θ) + 1)

arctan(x*10000pi)=126

arctan (x \cdot 10000 π) = 12 6^{\circ}

sin^2(x)*cos^2(x)=1

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

5.8=11.8sin(3.78*t)

5.8 = 11.8 sin (3.78 \cdot t)