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受欢迎的三角函数 >

sin^5(x)+2cos^2(x)=1

解答

sin^{5} (x) + 2 cos^{2} (x) = 1

解答

x = \frac{π}{2} + 2 πn, x = - 0.72214 \dots + 2 πn, x = π + 0.72214 \dots + 2 πn, x = 1.01290 \dots + 2 πn, x = π - 1.01290 \dots + 2 πn

+1

度数

x = 9 0^{\circ} + 36 0^{\circ} n, x = - 41.37561 \dots^{\circ} + 36 0^{\circ} n, x = 221.37561 \dots^{\circ} + 36 0^{\circ} n, x = 58.03535 \dots^{\circ} + 36 0^{\circ} n, x = 121.96464 \dots^{\circ} + 36 0^{\circ} n

求解步骤

sin^{5} (x) + 2 cos^{2} (x) = 1

两边减去

1

sin^{5} (x) + 2 cos^{2} (x) - 1 = 0

使用三角恒等式改写

- 1 + sin^{5} (x) + 2 cos^{2} (x)

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

= - 1 + sin^{5} (x) + 2 (1 - sin^{2} (x))

化简

- 1 + sin^{5} (x) + 2 (1 - sin^{2} (x)) : sin^{5} (x) - 2 sin^{2} (x) + 1

- 1 + sin^{5} (x) + 2 (1 - sin^{2} (x))

乘开

2 (1 - sin^{2} (x)) : 2 - 2 sin^{2} (x)

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

使用分配律:

a (b - c) = ab - a c

a = 2, b = 1, c = sin^{2} (x)

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

数字相乘:

2 \cdot 1 = 2

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

= - 1 + sin^{5} (x) + 2 - 2 sin^{2} (x)

化简

- 1 + sin^{5} (x) + 2 - 2 sin^{2} (x) : sin^{5} (x) - 2 sin^{2} (x) + 1

- 1 + sin^{5} (x) + 2 - 2 sin^{2} (x)

对同类项分组

= sin^{5} (x) - 2 sin^{2} (x) - 1 + 2

数字相加/相减:

- 1 + 2 = 1

= sin^{5} (x) - 2 sin^{2} (x) + 1

= sin^{5} (x) - 2 sin^{2} (x) + 1

= sin^{5} (x) - 2 sin^{2} (x) + 1

1 + sin^{5} (x) - 2 sin^{2} (x) = 0

用替代法求解

1 + sin^{5} (x) - 2 sin^{2} (x) = 0

令

: sin (x) = u

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

1 + u^{5} - 2 u^{2} = 0 : u = 1, u \approx - 0.66099 \dots, u \approx 0.84837 \dots

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

改写成标准形式

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

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

因式分解

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

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

使用有理根定理

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

a_{0}

的除数

: 1, a_{n}

的除数

: 1

因此，检验以下有理数:

\pm \frac{1}{1}

\frac{1}{1}

是表达式的根，所以因式分解

u - 1

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

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

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

对

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

做除法:

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

将分子

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

与除数

u - 1

的首项系数相除：

\frac{u ^{5}}{u} = u^{4}

商 = u^{4}

将

u - 1

乘以

u^{4} : u^{5} - u^{4}

将

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

减去

u^{5} - u^{4}

得到新的余数

余数 = u^{4} - 2 u^{2} + 1

因此

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

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

对

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

做除法:

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

将分子

u^{4} - 2 u^{2} + 1

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} - 2 u^{2} + 1

减去

u^{4} - u^{3}

得到新的余数

余数 = u^{3} - 2 u^{2} + 1

因此

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

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

对

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

做除法:

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

将分子

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

与除数

u - 1

的首项系数相除：

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

商 = u^{2}

将

u - 1

乘以

u^{2} : u^{3} - u^{2}

将

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

减去

u^{3} - u^{2}

得到新的余数

余数 = - u^{2} + 1

因此

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

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

对

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

做除法:

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

将分子

- u^{2} + 1

与除数

u - 1

的首项系数相除：

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

商 = - u

将

u - 1

乘以

- u : - u^{2} + u

将

- u^{2} + 1

减去

- u^{2} + u

得到新的余数

余数 = - u + 1

因此

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

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

对

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

做除法:

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

将分子

- u + 1

与除数

u - 1

的首项系数相除：

\frac{- u}{u} = - 1

商 = - 1

将

u - 1

乘以

- 1 : - u + 1

将

- u + 1

减去

- u + 1

得到新的余数

余数 = 0

因此

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

= u^{4} + u^{3} + u^{2} - u - 1

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

(u - 1) (u^{4} + u^{3} + u^{2} - u - 1) = 0

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

u - 1 = 0 or u^{4} + u^{3} + u^{2} - u - 1 = 0

解

u - 1 = 0 : u = 1

u - 1 = 0

将

1

到右边

u - 1 = 0

两边加上

1

u - 1 + 1 = 0 + 1

化简

u = 1

u = 1

解

u^{4} + u^{3} + u^{2} - u - 1 = 0 : u \approx - 0.66099 \dots, u \approx 0.84837 \dots

u^{4} + u^{3} + u^{2} - u - 1 = 0

使用牛顿-拉弗森方法找到

u^{4} + u^{3} + u^{2} - u - 1 = 0

的一个解:

u \approx - 0.66099 \dots

u^{4} + u^{3} + u^{2} - u - 1 = 0

牛顿-拉弗森近似法定义

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

找到

f^{^{'}} (u) : 4 u^{3} + 3 u^{2} + 2 u - 1

\frac{d}{d u} (u^{4} + u^{3} + u^{2} - u - 1)

使用微分加减法定则:

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

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

\frac{d}{d u} (u^{4}) = 4 u^{3}

\frac{d}{d u} (u^{4})

使用幂法则:

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

= 4 u^{4 - 1}

化简

= 4 u^{3}

\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 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} (1) = 0

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

常数微分:

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

= 0

= 4 u^{3} + 3 u^{2} + 2 u - 1 - 0

化简

= 4 u^{3} + 3 u^{2} + 2 u - 1

令

u_{0} = - 1

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = - 0.75 : Δ u_{1} = 0.25

f (u_{0}) = (- 1)^{4} + (- 1)^{3} + (- 1)^{2} - (- 1) - 1 = 1

f^{^{'}} (u_{0}) = 4 (- 1)^{3} + 3 (- 1)^{2} + 2 (- 1) - 1 = - 4

u_{1} = - 0.75

Δ u_{1} = ∣ - 0.75 - (- 1) ∣ = 0.25

Δ u_{1} = 0.25

u_{2} = - 0.6671875 : Δ u_{2} = 0.0828125

f (u_{1}) = (- 0.75)^{4} + (- 0.75)^{3} + (- 0.75)^{2} - (- 0.75) - 1 = 0.20703125

f^{^{'}} (u_{1}) = 4 (- 0.75)^{3} + 3 (- 0.75)^{2} + 2 (- 0.75) - 1 = - 2.5

u_{2} = - 0.6671875

Δ u_{2} = ∣ - 0.6671875 - (- 0.75) ∣ = 0.0828125

Δ u_{2} = 0.0828125

u_{3} = - 0.66102 \dots : Δ u_{3} = 0.00616 \dots

f (u_{2}) = (- 0.6671875)^{4} + (- 0.6671875)^{3} + (- 0.6671875)^{2} - (- 0.6671875) - 1 = 0.01348 \dots

f^{^{'}} (u_{2}) = 4 (- 0.6671875)^{3} + 3 (- 0.6671875)^{2} + 2 (- 0.6671875) - 1 = - 2.18692 \dots

u_{3} = - 0.66102 \dots

Δ u_{3} = ∣ - 0.66102 \dots - (- 0.6671875) ∣ = 0.00616 \dots

Δ u_{3} = 0.00616 \dots

u_{4} = - 0.66099 \dots : Δ u_{4} = 0.00002 \dots

f (u_{3}) = (- 0.66102 \dots)^{4} + (- 0.66102 \dots)^{3} + (- 0.66102 \dots)^{2} - (- 0.66102 \dots) - 1 = 0.00006 \dots

f^{^{'}} (u_{3}) = 4 (- 0.66102 \dots)^{3} + 3 (- 0.66102 \dots)^{2} + 2 (- 0.66102 \dots) - 1 = - 2.16652 \dots

u_{4} = - 0.66099 \dots

Δ u_{4} = ∣ - 0.66099 \dots - (- 0.66102 \dots) ∣ = 0.00002 \dots

Δ u_{4} = 0.00002 \dots

u_{5} = - 0.66099 \dots : Δ u_{5} = 6.41022 E - 10

f (u_{4}) = (- 0.66099 \dots)^{4} + (- 0.66099 \dots)^{3} + (- 0.66099 \dots)^{2} - (- 0.66099 \dots) - 1 = 1.38873 E - 9

f^{^{'}} (u_{4}) = 4 (- 0.66099 \dots)^{3} + 3 (- 0.66099 \dots)^{2} + 2 (- 0.66099 \dots) - 1 = - 2.16643 \dots

u_{5} = - 0.66099 \dots

Δ u_{5} = ∣ - 0.66099 \dots - (- 0.66099 \dots) ∣ = 6.41022 E - 10

Δ u_{5} = 6.41022 E - 10

u \approx - 0.66099 \dots

使用长除法

Eq u a t i o n 0 : \frac{u ^{4} + u ^{3} + u ^{2} - u - 1}{u + 0.66099 \dots} = u^{3} + 0.33900 \dots u^{2} + 0.77591 \dots u - 1.51287 \dots

u^{3} + 0.33900 \dots u^{2} + 0.77591 \dots u - 1.51287 \dots \approx 0

使用牛顿-拉弗森方法找到

u^{3} + 0.33900 \dots u^{2} + 0.77591 \dots u - 1.51287 \dots = 0

的一个解:

u \approx 0.84837 \dots

u^{3} + 0.33900 \dots u^{2} + 0.77591 \dots u - 1.51287 \dots = 0

牛顿-拉弗森近似法定义

f (u) = u^{3} + 0.33900 \dots u^{2} + 0.77591 \dots u - 1.51287 \dots

找到

f^{^{'}} (u) : 3 u^{2} + 0.67801 \dots u + 0.77591 \dots

\frac{d}{d u} (u^{3} + 0.33900 \dots u^{2} + 0.77591 \dots u - 1.51287 \dots)

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} (0.33900 \dots u^{2}) + \frac{d}{d u} (0.77591 \dots u) - \frac{d}{d u} (1.51287 \dots)

\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} (0.33900 \dots u^{2}) = 0.67801 \dots u

\frac{d}{d u} (0.33900 \dots u^{2})

将常数提出:

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

= 0.33900 \dots \frac{d}{d u} (u^{2})

使用幂法则:

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

= 0.33900 \dots \cdot 2 u^{2 - 1}

化简

= 0.67801 \dots u

\frac{d}{d u} (0.77591 \dots u) = 0.77591 \dots

\frac{d}{d u} (0.77591 \dots u)

将常数提出:

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

= 0.77591 \dots \frac{d u}{d u}

使用常见微分定则:

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

= 0.77591 \dots \cdot 1

化简

= 0.77591 \dots

\frac{d}{d u} (1.51287 \dots) = 0

\frac{d}{d u} (1.51287 \dots)

常数微分:

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

= 0

= 3 u^{2} + 0.67801 \dots u + 0.77591 \dots - 0

化简

= 3 u^{2} + 0.67801 \dots u + 0.77591 \dots

令

u_{0} = 2

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 1.33519 \dots : Δ u_{1} = 0.66480 \dots

f (u_{0}) = 2^{3} + 0.33900 \dots \cdot 2^{2} + 0.77591 \dots \cdot 2 - 1.51287 \dots = 9.39499 \dots

f^{^{'}} (u_{0}) = 3 \cdot 2^{2} + 0.67801 \dots \cdot 2 + 0.77591 \dots = 14.13194 \dots

u_{1} = 1.33519 \dots

Δ u_{1} = ∣ 1.33519 \dots - 2 ∣ = 0.66480 \dots

Δ u_{1} = 0.66480 \dots

u_{2} = 0.97843 \dots : Δ u_{2} = 0.35675 \dots

f (u_{1}) = 1.33519 \dots^{3} + 0.33900 \dots \cdot 1.33519 \dots^{2} + 0.77591 \dots \cdot 1.33519 \dots - 1.51287 \dots = 2.50780 \dots

f^{^{'}} (u_{1}) = 3 \cdot 1.33519 \dots^{2} + 0.67801 \dots \cdot 1.33519 \dots + 0.77591 \dots = 7.02943 \dots

u_{2} = 0.97843 \dots

Δ u_{2} = ∣ 0.97843 \dots - 1.33519 \dots ∣ = 0.35675 \dots

Δ u_{2} = 0.35675 \dots

u_{3} = 0.86071 \dots : Δ u_{3} = 0.11772 \dots

f (u_{2}) = 0.97843 \dots^{3} + 0.33900 \dots \cdot 0.97843 \dots^{2} + 0.77591 \dots \cdot 0.97843 \dots - 1.51287 \dots = 0.50755 \dots

f^{^{'}} (u_{2}) = 3 \cdot 0.97843 \dots^{2} + 0.67801 \dots \cdot 0.97843 \dots + 0.77591 \dots = 4.31133 \dots

u_{3} = 0.86071 \dots

Δ u_{3} = ∣ 0.86071 \dots - 0.97843 \dots ∣ = 0.11772 \dots

Δ u_{3} = 0.11772 \dots

u_{4} = 0.84849 \dots : Δ u_{4} = 0.01221 \dots

f (u_{3}) = 0.86071 \dots^{3} + 0.33900 \dots \cdot 0.86071 \dots^{2} + 0.77591 \dots \cdot 0.86071 \dots - 1.51287 \dots = 0.04374 \dots

f^{^{'}} (u_{3}) = 3 \cdot 0.86071 \dots^{2} + 0.67801 \dots \cdot 0.86071 \dots + 0.77591 \dots = 3.58196 \dots

u_{4} = 0.84849 \dots

Δ u_{4} = ∣ 0.84849 \dots - 0.86071 \dots ∣ = 0.01221 \dots

Δ u_{4} = 0.01221 \dots

u_{5} = 0.84837 \dots : Δ u_{5} = 0.00012 \dots

f (u_{4}) = 0.84849 \dots^{3} + 0.33900 \dots \cdot 0.84849 \dots^{2} + 0.77591 \dots \cdot 0.84849 \dots - 1.51287 \dots = 0.00043 \dots

f^{^{'}} (u_{4}) = 3 \cdot 0.84849 \dots^{2} + 0.67801 \dots \cdot 0.84849 \dots + 0.77591 \dots = 3.51106 \dots

u_{5} = 0.84837 \dots

Δ u_{5} = ∣ 0.84837 \dots - 0.84849 \dots ∣ = 0.00012 \dots

Δ u_{5} = 0.00012 \dots

u_{6} = 0.84837 \dots : Δ u_{6} = 1.25501 E - 8

f (u_{5}) = 0.84837 \dots^{3} + 0.33900 \dots \cdot 0.84837 \dots^{2} + 0.77591 \dots \cdot 0.84837 \dots - 1.51287 \dots = 4.40554 E - 8

f^{^{'}} (u_{5}) = 3 \cdot 0.84837 \dots^{2} + 0.67801 \dots \cdot 0.84837 \dots + 0.77591 \dots = 3.51034 \dots

u_{6} = 0.84837 \dots

Δ u_{6} = ∣ 0.84837 \dots - 0.84837 \dots ∣ = 1.25501 E - 8

Δ u_{6} = 1.25501 E - 8

u \approx 0.84837 \dots

使用长除法

Eq u a t i o n 0 : \frac{u ^{3} + 0.33900 \dots u ^{2} + 0.77591 \dots u - 1.51287 \dots}{u - 0.84837 \dots} = u^{2} + 1.18738 \dots u + 1.78326 \dots

u^{2} + 1.18738 \dots u + 1.78326 \dots \approx 0

使用牛顿-拉弗森方法找到

u^{2} + 1.18738 \dots u + 1.78326 \dots = 0

的一个解:

u \in R

无解

u^{2} + 1.18738 \dots u + 1.78326 \dots = 0

牛顿-拉弗森近似法定义

f (u) = u^{2} + 1.18738 \dots u + 1.78326 \dots

找到

f^{^{'}} (u) : 2 u + 1.18738 \dots

\frac{d}{d u} (u^{2} + 1.18738 \dots u + 1.78326 \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} (1.18738 \dots u) + \frac{d}{d u} (1.78326 \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} (1.18738 \dots u) = 1.18738 \dots

\frac{d}{d u} (1.18738 \dots u)

将常数提出:

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

= 1.18738 \dots \frac{d u}{d u}

使用常见微分定则:

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

= 1.18738 \dots \cdot 1

化简

= 1.18738 \dots

\frac{d}{d u} (1.78326 \dots) = 0

\frac{d}{d u} (1.78326 \dots)

常数微分:

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

= 0

= 2 u + 1.18738 \dots + 0

化简

= 2 u + 1.18738 \dots

令

u_{0} = - 2

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = - 0.78813 \dots : Δ u_{1} = 1.21186 \dots

f (u_{0}) = (- 2)^{2} + 1.18738 \dots (- 2) + 1.78326 \dots = 3.40849 \dots

f^{^{'}} (u_{0}) = 2 (- 2) + 1.18738 \dots = - 2.81261 \dots

u_{1} = - 0.78813 \dots

Δ u_{1} = ∣ - 0.78813 \dots - (- 2) ∣ = 1.21186 \dots

Δ u_{1} = 1.21186 \dots

u_{2} = 2.98819 \dots : Δ u_{2} = 3.77633 \dots

f (u_{1}) = (- 0.78813 \dots)^{2} + 1.18738 \dots (- 0.78813 \dots) + 1.78326 \dots = 1.46860 \dots

f^{^{'}} (u_{1}) = 2 (- 0.78813 \dots) + 1.18738 \dots = - 0.38889 \dots

u_{2} = 2.98819 \dots

Δ u_{2} = ∣ 2.98819 \dots - (- 0.78813 \dots) ∣ = 3.77633 \dots

Δ u_{2} = 3.77633 \dots

u_{3} = 0.99752 \dots : Δ u_{3} = 1.99066 \dots

f (u_{2}) = 2.98819 \dots^{2} + 1.18738 \dots \cdot 2.98819 \dots + 1.78326 \dots = 14.26067 \dots

f^{^{'}} (u_{2}) = 2 \cdot 2.98819 \dots + 1.18738 \dots = 7.16376 \dots

u_{3} = 0.99752 \dots

Δ u_{3} = ∣ 0.99752 \dots - 2.98819 \dots ∣ = 1.99066 \dots

Δ u_{3} = 1.99066 \dots

u_{4} = - 0.24767 \dots : Δ u_{4} = 1.24519 \dots

f (u_{3}) = 0.99752 \dots^{2} + 1.18738 \dots \cdot 0.99752 \dots + 1.78326 \dots = 3.96275 \dots

f^{^{'}} (u_{3}) = 2 \cdot 0.99752 \dots + 1.18738 \dots = 3.18242 \dots

u_{4} = - 0.24767 \dots

Δ u_{4} = ∣ - 0.24767 \dots - 0.99752 \dots ∣ = 1.24519 \dots

Δ u_{4} = 1.24519 \dots

u_{5} = - 2.48821 \dots : Δ u_{5} = 2.24053 \dots

f (u_{4}) = (- 0.24767 \dots)^{2} + 1.18738 \dots (- 0.24767 \dots) + 1.78326 \dots = 1.55052 \dots

f^{^{'}} (u_{4}) = 2 (- 0.24767 \dots) + 1.18738 \dots = 0.69203 \dots

u_{5} = - 2.48821 \dots

Δ u_{5} = ∣ - 2.48821 \dots - (- 0.24767 \dots) ∣ = 2.24053 \dots

Δ u_{5} = 2.24053 \dots

u_{6} = - 1.16333 \dots : Δ u_{6} = 1.32487 \dots

f (u_{5}) = (- 2.48821 \dots)^{2} + 1.18738 \dots (- 2.48821 \dots) + 1.78326 \dots = 5.02001 \dots

f^{^{'}} (u_{5}) = 2 (- 2.48821 \dots) + 1.18738 \dots = - 3.78904 \dots

u_{6} = - 1.16333 \dots

Δ u_{6} = ∣ - 1.16333 \dots - (- 2.48821 \dots) ∣ = 1.32487 \dots

Δ u_{6} = 1.32487 \dots

u_{7} = 0.37734 \dots : Δ u_{7} = 1.54068 \dots

f (u_{6}) = (- 1.16333 \dots)^{2} + 1.18738 \dots (- 1.16333 \dots) + 1.78326 \dots = 1.75529 \dots

f^{^{'}} (u_{6}) = 2 (- 1.16333 \dots) + 1.18738 \dots = - 1.13929 \dots

u_{7} = 0.37734 \dots

Δ u_{7} = ∣ 0.37734 \dots - (- 1.16333 \dots) ∣ = 1.54068 \dots

Δ u_{7} = 1.54068 \dots

u_{8} = - 0.84491 \dots : Δ u_{8} = 1.22225 \dots

f (u_{7}) = 0.37734 \dots^{2} + 1.18738 \dots \cdot 0.37734 \dots + 1.78326 \dots = 2.37370 \dots

f^{^{'}} (u_{7}) = 2 \cdot 0.37734 \dots + 1.18738 \dots = 1.94206 \dots

u_{8} = - 0.84491 \dots

Δ u_{8} = ∣ - 0.84491 \dots - 0.37734 \dots ∣ = 1.22225 \dots

Δ u_{8} = 1.22225 \dots

u_{9} = 2.12837 \dots : Δ u_{9} = 2.97328 \dots

f (u_{8}) = (- 0.84491 \dots)^{2} + 1.18738 \dots (- 0.84491 \dots) + 1.78326 \dots = 1.49390 \dots

f^{^{'}} (u_{8}) = 2 (- 0.84491 \dots) + 1.18738 \dots = - 0.50244 \dots

u_{9} = 2.12837 \dots

Δ u_{9} = ∣ 2.12837 \dots - (- 0.84491 \dots) ∣ = 2.97328 \dots

Δ u_{9} = 2.97328 \dots

无法得出解

解为

u \approx - 0.66099 \dots, u \approx 0.84837 \dots

解为

u = 1, u \approx - 0.66099 \dots, u \approx 0.84837 \dots

u = sin (x)

代回

sin (x) = 1, sin (x) \approx - 0.66099 \dots, sin (x) \approx 0.84837 \dots

sin (x) = 1, sin (x) \approx - 0.66099 \dots, sin (x) \approx 0.84837 \dots

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

sin (x) = 1

的通解

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

sin (x) = - 0.66099 \dots : x = arcsin (- 0.66099 \dots) + 2 πn, x = π + arcsin (0.66099 \dots) + 2 πn

sin (x) = - 0.66099 \dots

使用反三角函数性质

sin (x) = - 0.66099 \dots

sin (x) = - 0.66099 \dots

的通解

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- 0.66099 \dots) + 2 πn, x = π + arcsin (0.66099 \dots) + 2 πn

x = arcsin (- 0.66099 \dots) + 2 πn, x = π + arcsin (0.66099 \dots) + 2 πn

sin (x) = 0.84837 \dots : x = arcsin (0.84837 \dots) + 2 πn, x = π - arcsin (0.84837 \dots) + 2 πn

sin (x) = 0.84837 \dots

使用反三角函数性质

sin (x) = 0.84837 \dots

sin (x) = 0.84837 \dots

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (0.84837 \dots) + 2 πn, x = π - arcsin (0.84837 \dots) + 2 πn

x = arcsin (0.84837 \dots) + 2 πn, x = π - arcsin (0.84837 \dots) + 2 πn

合并所有解

x = \frac{π}{2} + 2 πn, x = arcsin (- 0.66099 \dots) + 2 πn, x = π + arcsin (0.66099 \dots) + 2 πn, x = arcsin (0.84837 \dots) + 2 πn, x = π - arcsin (0.84837 \dots) + 2 πn

以小数形式表示解

x = \frac{π}{2} + 2 πn, x = - 0.72214 \dots + 2 πn, x = π + 0.72214 \dots + 2 πn, x = 1.01290 \dots + 2 πn, x = π - 1.01290 \dots + 2 πn

作图

流行的例子

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