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受欢迎的三角函数 >

cos^5(x)+cos(x)+4cos^2(x)=2

解答

cos^{5} (x) + cos (x) + 4 cos^{2} (x) = 2

解答

x = π + 2 πn, x = 0.95033 \dots + 2 πn, x = 2 π - 0.95033 \dots + 2 πn

+1

度数

x = 18 0^{\circ} + 36 0^{\circ} n, x = 54.45019 \dots^{\circ} + 36 0^{\circ} n, x = 305.54980 \dots^{\circ} + 36 0^{\circ} n

求解步骤

cos^{5} (x) + cos (x) + 4 cos^{2} (x) = 2

用替代法求解

cos^{5} (x) + cos (x) + 4 cos^{2} (x) = 2

令

: cos (x) = u

u^{5} + u + 4 u^{2} = 2

u^{5} + u + 4 u^{2} = 2 : u = - 1, u \approx 0.58141 \dots, u \approx - 1.23238 \dots

u^{5} + u + 4 u^{2} = 2

将

2

para o lado esquerdo

u^{5} + u + 4 u^{2} = 2

两边减去

2

u^{5} + u + 4 u^{2} - 2 = 2 - 2

化简

u^{5} + u + 4 u^{2} - 2 = 0

u^{5} + u + 4 u^{2} - 2 = 0

改写成标准形式

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

u^{5} + 4 u^{2} + u - 2 = 0

因式分解

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

u^{5} + 4 u^{2} + u - 2

使用有理根定理

a_{0} = 2, a_{n} = 1

a_{0}

的除数

: 1, 2, a_{n}

的除数

: 1

因此，检验以下有理数:

\pm \frac{1 , 2}{1}

- \frac{1}{1}

是表达式的根，所以因式分解

u + 1

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

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

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

对

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

做除法:

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

将分子

u^{5} + 4 u^{2} + u - 2

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

减去

u^{5} + u^{4}

得到新的余数

余数 = - u^{4} + 4 u^{2} + u - 2

因此

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

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

对

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

做除法:

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

将分子

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

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

减去

- u^{4} - u^{3}

得到新的余数

余数 = u^{3} + 4 u^{2} + u - 2

因此

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

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

对

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

做除法:

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

将分子

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

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

减去

u^{3} + u^{2}

得到新的余数

余数 = 3 u^{2} + u - 2

因此

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

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

对

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

做除法:

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

将分子

3 u^{2} + u - 2

与除数

u + 1

的首项系数相除：

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

商 = 3 u

将

u + 1

乘以

3 u : 3 u^{2} + 3 u

将

3 u^{2} + u - 2

减去

3 u^{2} + 3 u

得到新的余数

余数 = - 2 u - 2

因此

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

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

对

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

做除法:

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

将分子

- 2 u - 2

与除数

u + 1

的首项系数相除：

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

商 = - 2

将

u + 1

乘以

- 2 : - 2 u - 2

将

- 2 u - 2

减去

- 2 u - 2

得到新的余数

余数 = 0

因此

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

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

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

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

解

u + 1 = 0 : u = - 1

u + 1 = 0

将

1

到右边

u + 1 = 0

两边减去

1

u + 1 - 1 = 0 - 1

化简

u = - 1

u = - 1

解

u^{4} - u^{3} + u^{2} + 3 u - 2 = 0 : u \approx 0.58141 \dots, u \approx - 1.23238 \dots

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

使用牛顿-拉弗森方法找到

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

的一个解:

u \approx 0.58141 \dots

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

牛顿-拉弗森近似法定义

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

找到

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

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

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}{d u} (3 u) - \frac{d}{d u} (2)

\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}{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} (2) = 0

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

常数微分:

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

= 0

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

化简

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

令

u_{0} = 1

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 0.66666 \dots : Δ u_{1} = 0.33333 \dots

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

f^{^{'}} (u_{0}) = 4 \cdot 1^{3} - 3 \cdot 1^{2} + 2 \cdot 1 + 3 = 6

u_{1} = 0.66666 \dots

Δ u_{1} = ∣ 0.66666 \dots - 1 ∣ = 0.33333 \dots

Δ u_{1} = 0.33333 \dots

u_{2} = 0.58407 \dots : Δ u_{2} = 0.08259 \dots

f (u_{1}) = 0.66666 \dots^{4} - 0.66666 \dots^{3} + 0.66666 \dots^{2} + 3 \cdot 0.66666 \dots - 2 = 0.34567 \dots

f^{^{'}} (u_{1}) = 4 \cdot 0.66666 \dots^{3} - 3 \cdot 0.66666 \dots^{2} + 2 \cdot 0.66666 \dots + 3 = 4.18518 \dots

u_{2} = 0.58407 \dots

Δ u_{2} = ∣ 0.58407 \dots - 0.66666 \dots ∣ = 0.08259 \dots

Δ u_{2} = 0.08259 \dots

u_{3} = 0.58141 \dots : Δ u_{3} = 0.00265 \dots

f (u_{2}) = 0.58407 \dots^{4} - 0.58407 \dots^{3} + 0.58407 \dots^{2} + 3 \cdot 0.58407 \dots - 2 = 0.01047 \dots

f^{^{'}} (u_{2}) = 4 \cdot 0.58407 \dots^{3} - 3 \cdot 0.58407 \dots^{2} + 2 \cdot 0.58407 \dots + 3 = 3.94172 \dots

u_{3} = 0.58141 \dots

Δ u_{3} = ∣ 0.58141 \dots - 0.58407 \dots ∣ = 0.00265 \dots

Δ u_{3} = 0.00265 \dots

u_{4} = 0.58141 \dots : Δ u_{4} = 2.31829 E - 6

f (u_{3}) = 0.58141 \dots^{4} - 0.58141 \dots^{3} + 0.58141 \dots^{2} + 3 \cdot 0.58141 \dots - 2 = 9.12218 E - 6

f^{^{'}} (u_{3}) = 4 \cdot 0.58141 \dots^{3} - 3 \cdot 0.58141 \dots^{2} + 2 \cdot 0.58141 \dots + 3 = 3.93486 \dots

u_{4} = 0.58141 \dots

Δ u_{4} = ∣ 0.58141 \dots - 0.58141 \dots ∣ = 2.31829 E - 6

Δ u_{4} = 2.31829 E - 6

u_{5} = 0.58141 \dots : Δ u_{5} = 1.75373 E - 12

f (u_{4}) = 0.58141 \dots^{4} - 0.58141 \dots^{3} + 0.58141 \dots^{2} + 3 \cdot 0.58141 \dots - 2 = 6.9007 E - 12

f^{^{'}} (u_{4}) = 4 \cdot 0.58141 \dots^{3} - 3 \cdot 0.58141 \dots^{2} + 2 \cdot 0.58141 \dots + 3 = 3.93486 \dots

u_{5} = 0.58141 \dots

Δ u_{5} = ∣ 0.58141 \dots - 0.58141 \dots ∣ = 1.75373 E - 12

Δ u_{5} = 1.75373 E - 12

u \approx 0.58141 \dots

使用长除法

Eq u a t i o n 0 : \frac{u ^{4} - u ^{3} + u ^{2} + 3 u - 2}{u - 0.58141 \dots} = u^{3} - 0.41858 \dots u^{2} + 0.75662 \dots u + 3.43991 \dots

u^{3} - 0.41858 \dots u^{2} + 0.75662 \dots u + 3.43991 \dots \approx 0

使用牛顿-拉弗森方法找到

u^{3} - 0.41858 \dots u^{2} + 0.75662 \dots u + 3.43991 \dots = 0

的一个解:

u \approx - 1.23238 \dots

u^{3} - 0.41858 \dots u^{2} + 0.75662 \dots u + 3.43991 \dots = 0

牛顿-拉弗森近似法定义

f (u) = u^{3} - 0.41858 \dots u^{2} + 0.75662 \dots u + 3.43991 \dots

找到

f^{^{'}} (u) : 3 u^{2} - 0.83717 \dots u + 0.75662 \dots

\frac{d}{d u} (u^{3} - 0.41858 \dots u^{2} + 0.75662 \dots u + 3.43991 \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} (0.41858 \dots u^{2}) + \frac{d}{d u} (0.75662 \dots u) + \frac{d}{d u} (3.43991 \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} (0.41858 \dots u^{2}) = 0.83717 \dots u

\frac{d}{d u} (0.41858 \dots u^{2})

将常数提出:

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

= 0.41858 \dots \frac{d}{d u} (u^{2})

使用幂法则:

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

= 0.41858 \dots \cdot 2 u^{2 - 1}

化简

= 0.83717 \dots u

\frac{d}{d u} (0.75662 \dots u) = 0.75662 \dots

\frac{d}{d u} (0.75662 \dots u)

将常数提出:

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

= 0.75662 \dots \frac{d u}{d u}

使用常见微分定则:

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

= 0.75662 \dots \cdot 1

化简

= 0.75662 \dots

\frac{d}{d u} (3.43991 \dots) = 0

\frac{d}{d u} (3.43991 \dots)

常数微分:

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

= 0

= 3 u^{2} - 0.83717 \dots u + 0.75662 \dots + 0

化简

= 3 u^{2} - 0.83717 \dots u + 0.75662 \dots

令

u_{0} = - 5

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = - 3.30117 \dots : Δ u_{1} = 1.69882 \dots

f (u_{0}) = (- 5)^{3} - 0.41858 \dots (- 5)^{2} + 0.75662 \dots (- 5) + 3.43991 \dots = - 135.80796 \dots

f^{^{'}} (u_{0}) = 3 (- 5)^{2} - 0.83717 \dots (- 5) + 0.75662 \dots = 79.94252 \dots

u_{1} = - 3.30117 \dots

Δ u_{1} = ∣ - 3.30117 \dots - (- 5) ∣ = 1.69882 \dots

Δ u_{1} = 1.69882 \dots

u_{2} = - 2.20780 \dots : Δ u_{2} = 1.09337 \dots

f (u_{1}) = (- 3.30117 \dots)^{3} - 0.41858 \dots (- 3.30117 \dots)^{2} + 0.75662 \dots (- 3.30117 \dots) + 3.43991 \dots = - 39.59511 \dots

f^{^{'}} (u_{1}) = 3 (- 3.30117 \dots)^{2} - 0.83717 \dots (- 3.30117 \dots) + 0.75662 \dots = 36.21367 \dots

u_{2} = - 2.20780 \dots

Δ u_{2} = ∣ - 2.20780 \dots - (- 3.30117 \dots) ∣ = 1.09337 \dots

Δ u_{2} = 1.09337 \dots

u_{3} = - 1.56741 \dots : Δ u_{3} = 0.64038 \dots

f (u_{2}) = (- 2.20780 \dots)^{3} - 0.41858 \dots (- 2.20780 \dots)^{2} + 0.75662 \dots (- 2.20780 \dots) + 3.43991 \dots = - 11.03268 \dots

f^{^{'}} (u_{2}) = 3 (- 2.20780 \dots)^{2} - 0.83717 \dots (- 2.20780 \dots) + 0.75662 \dots = 17.22816 \dots

u_{3} = - 1.56741 \dots

Δ u_{3} = ∣ - 1.56741 \dots - (- 2.20780 \dots) ∣ = 0.64038 \dots

Δ u_{3} = 0.64038 \dots

u_{4} = - 1.28929 \dots : Δ u_{4} = 0.27812 \dots

f (u_{3}) = (- 1.56741 \dots)^{3} - 0.41858 \dots (- 1.56741 \dots)^{2} + 0.75662 \dots (- 1.56741 \dots) + 3.43991 \dots = - 2.62526 \dots

f^{^{'}} (u_{3}) = 3 (- 1.56741 \dots)^{2} - 0.83717 \dots (- 1.56741 \dots) + 0.75662 \dots = 9.43924 \dots

u_{4} = - 1.28929 \dots

Δ u_{4} = ∣ - 1.28929 \dots - (- 1.56741 \dots) ∣ = 0.27812 \dots

Δ u_{4} = 0.27812 \dots

u_{5} = - 1.23439 \dots : Δ u_{5} = 0.05490 \dots

f (u_{4}) = (- 1.28929 \dots)^{3} - 0.41858 \dots (- 1.28929 \dots)^{2} + 0.75662 \dots (- 1.28929 \dots) + 3.43991 \dots = - 0.37459 \dots

f^{^{'}} (u_{4}) = 3 (- 1.28929 \dots)^{2} - 0.83717 \dots (- 1.28929 \dots) + 0.75662 \dots = 6.82285 \dots

u_{5} = - 1.23439 \dots

Δ u_{5} = ∣ - 1.23439 \dots - (- 1.28929 \dots) ∣ = 0.05490 \dots

Δ u_{5} = 0.05490 \dots

u_{6} = - 1.23238 \dots : Δ u_{6} = 0.00200 \dots

f (u_{5}) = (- 1.23439 \dots)^{3} - 0.41858 \dots (- 1.23439 \dots)^{2} + 0.75662 \dots (- 1.23439 \dots) + 3.43991 \dots = - 0.01275 \dots

f^{^{'}} (u_{5}) = 3 (- 1.23439 \dots)^{2} - 0.83717 \dots (- 1.23439 \dots) + 0.75662 \dots = 6.36121 \dots

u_{6} = - 1.23238 \dots

Δ u_{6} = ∣ - 1.23238 \dots - (- 1.23439 \dots) ∣ = 0.00200 \dots

Δ u_{6} = 0.00200 \dots

u_{7} = - 1.23238 \dots : Δ u_{7} = 2.61081 E - 6

f (u_{6}) = (- 1.23238 \dots)^{3} - 0.41858 \dots (- 1.23238 \dots)^{2} + 0.75662 \dots (- 1.23238 \dots) + 3.43991 \dots = - 0.00001 \dots

f^{^{'}} (u_{6}) = 3 (- 1.23238 \dots)^{2} - 0.83717 \dots (- 1.23238 \dots) + 0.75662 \dots = 6.34469 \dots

u_{7} = - 1.23238 \dots

Δ u_{7} = ∣ - 1.23238 \dots - (- 1.23238 \dots) ∣ = 2.61081 E - 6

Δ u_{7} = 2.61081 E - 6

u_{8} = - 1.23238 \dots : Δ u_{8} = 4.42173 E - 12

f (u_{7}) = (- 1.23238 \dots)^{3} - 0.41858 \dots (- 1.23238 \dots)^{2} + 0.75662 \dots (- 1.23238 \dots) + 3.43991 \dots = - 2.80544 E - 11

f^{^{'}} (u_{7}) = 3 (- 1.23238 \dots)^{2} - 0.83717 \dots (- 1.23238 \dots) + 0.75662 \dots = 6.34467 \dots

u_{8} = - 1.23238 \dots

Δ u_{8} = ∣ - 1.23238 \dots - (- 1.23238 \dots) ∣ = 4.42173 E - 12

Δ u_{8} = 4.42173 E - 12

u \approx - 1.23238 \dots

使用长除法

Eq u a t i o n 0 : \frac{u ^{3} - 0.41858 \dots u ^{2} + 0.75662 \dots u + 3.43991 \dots}{u + 1.23238 \dots} = u^{2} - 1.65097 \dots u + 2.79126 \dots

u^{2} - 1.65097 \dots u + 2.79126 \dots \approx 0

使用牛顿-拉弗森方法找到

u^{2} - 1.65097 \dots u + 2.79126 \dots = 0

的一个解:

u \in R

无解

u^{2} - 1.65097 \dots u + 2.79126 \dots = 0

牛顿-拉弗森近似法定义

f (u) = u^{2} - 1.65097 \dots u + 2.79126 \dots

找到

f^{^{'}} (u) : 2 u - 1.65097 \dots

\frac{d}{d u} (u^{2} - 1.65097 \dots u + 2.79126 \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} (1.65097 \dots u) + \frac{d}{d u} (2.79126 \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} (1.65097 \dots u) = 1.65097 \dots

\frac{d}{d u} (1.65097 \dots u)

将常数提出:

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

= 1.65097 \dots \frac{d u}{d u}

使用常见微分定则:

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

= 1.65097 \dots \cdot 1

化简

= 1.65097 \dots

\frac{d}{d u} (2.79126 \dots) = 0

\frac{d}{d u} (2.79126 \dots)

常数微分:

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

= 0

= 2 u - 1.65097 \dots + 0

化简

= 2 u - 1.65097 \dots

令

u_{0} = 2

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 0.51456 \dots : Δ u_{1} = 1.48543 \dots

f (u_{0}) = 2^{2} - 1.65097 \dots \cdot 2 + 2.79126 \dots = 3.48931 \dots

f^{^{'}} (u_{0}) = 2 \cdot 2 - 1.65097 \dots = 2.34902 \dots

u_{1} = 0.51456 \dots

Δ u_{1} = ∣ 0.51456 \dots - 2 ∣ = 1.48543 \dots

Δ u_{1} = 1.48543 \dots

u_{2} = 4.06293 \dots : Δ u_{2} = 3.54837 \dots

f (u_{1}) = 0.51456 \dots^{2} - 1.65097 \dots \cdot 0.51456 \dots + 2.79126 \dots = 2.20650 \dots

f^{^{'}} (u_{1}) = 2 \cdot 0.51456 \dots - 1.65097 \dots = - 0.62183 \dots

u_{2} = 4.06293 \dots

Δ u_{2} = ∣ 4.06293 \dots - 0.51456 \dots ∣ = 3.54837 \dots

Δ u_{2} = 3.54837 \dots

u_{3} = 2.11836 \dots : Δ u_{3} = 1.94457 \dots

f (u_{2}) = 4.06293 \dots^{2} - 1.65097 \dots \cdot 4.06293 \dots + 2.79126 \dots = 12.59093 \dots

f^{^{'}} (u_{2}) = 2 \cdot 4.06293 \dots - 1.65097 \dots = 6.47490 \dots

u_{3} = 2.11836 \dots

Δ u_{3} = ∣ 2.11836 \dots - 4.06293 \dots ∣ = 1.94457 \dots

Δ u_{3} = 1.94457 \dots

u_{4} = 0.65598 \dots : Δ u_{4} = 1.46238 \dots

f (u_{3}) = 2.11836 \dots^{2} - 1.65097 \dots \cdot 2.11836 \dots + 2.79126 \dots = 3.78136 \dots

f^{^{'}} (u_{3}) = 2 \cdot 2.11836 \dots - 1.65097 \dots = 2.58575 \dots

u_{4} = 0.65598 \dots

Δ u_{4} = ∣ 0.65598 \dots - 2.11836 \dots ∣ = 1.46238 \dots

Δ u_{4} = 1.46238 \dots

u_{5} = 6.96423 \dots : Δ u_{5} = 6.30824 \dots

f (u_{4}) = 0.65598 \dots^{2} - 1.65097 \dots \cdot 0.65598 \dots + 2.79126 \dots = 2.13856 \dots

f^{^{'}} (u_{4}) = 2 \cdot 0.65598 \dots - 1.65097 \dots = - 0.33901 \dots

u_{5} = 6.96423 \dots

Δ u_{5} = ∣ 6.96423 \dots - 0.65598 \dots ∣ = 6.30824 \dots

Δ u_{5} = 6.30824 \dots

u_{6} = 3.72301 \dots : Δ u_{6} = 3.24121 \dots

f (u_{5}) = 6.96423 \dots^{2} - 1.65097 \dots \cdot 6.96423 \dots + 2.79126 \dots = 39.79400 \dots

f^{^{'}} (u_{5}) = 2 \cdot 6.96423 \dots - 1.65097 \dots = 12.27748 \dots

u_{6} = 3.72301 \dots

Δ u_{6} = ∣ 3.72301 \dots - 6.96423 \dots ∣ = 3.24121 \dots

Δ u_{6} = 3.24121 \dots

u_{7} = 1.91017 \dots : Δ u_{7} = 1.81283 \dots

f (u_{6}) = 3.72301 \dots^{2} - 1.65097 \dots \cdot 3.72301 \dots + 2.79126 \dots = 10.50549 \dots

f^{^{'}} (u_{6}) = 2 \cdot 3.72301 \dots - 1.65097 \dots = 5.79505 \dots

u_{7} = 1.91017 \dots

Δ u_{7} = ∣ 1.91017 \dots - 3.72301 \dots ∣ = 1.81283 \dots

Δ u_{7} = 1.81283 \dots

u_{8} = 0.39527 \dots : Δ u_{8} = 1.51489 \dots

f (u_{7}) = 1.91017 \dots^{2} - 1.65097 \dots \cdot 1.91017 \dots + 2.79126 \dots = 3.28638 \dots

f^{^{'}} (u_{7}) = 2 \cdot 1.91017 \dots - 1.65097 \dots = 2.16937 \dots

u_{8} = 0.39527 \dots

Δ u_{8} = ∣ 0.39527 \dots - 1.91017 \dots ∣ = 1.51489 \dots

Δ u_{8} = 1.51489 \dots

u_{9} = 3.06248 \dots : Δ u_{9} = 2.66720 \dots

f (u_{8}) = 0.39527 \dots^{2} - 1.65097 \dots \cdot 0.39527 \dots + 2.79126 \dots = 2.29491 \dots

f^{^{'}} (u_{8}) = 2 \cdot 0.39527 \dots - 1.65097 \dots = - 0.86041 \dots

u_{9} = 3.06248 \dots

Δ u_{9} = ∣ 3.06248 \dots - 0.39527 \dots ∣ = 2.66720 \dots

Δ u_{9} = 2.66720 \dots

u_{10} = 1.47240 \dots : Δ u_{10} = 1.59007 \dots

f (u_{9}) = 3.06248 \dots^{2} - 1.65097 \dots \cdot 3.06248 \dots + 2.79126 \dots = 7.11398 \dots

f^{^{'}} (u_{9}) = 2 \cdot 3.06248 \dots - 1.65097 \dots = 4.47398 \dots

u_{10} = 1.47240 \dots

Δ u_{10} = ∣ 1.47240 \dots - 3.06248 \dots ∣ = 1.59007 \dots

Δ u_{10} = 1.59007 \dots

u_{11} = - 0.48173 \dots : Δ u_{11} = 1.95413 \dots

f (u_{10}) = 1.47240 \dots^{2} - 1.65097 \dots \cdot 1.47240 \dots + 2.79126 \dots = 2.52833 \dots

f^{^{'}} (u_{10}) = 2 \cdot 1.47240 \dots - 1.65097 \dots = 1.29383 \dots

u_{11} = - 0.48173 \dots

Δ u_{11} = ∣ - 0.48173 \dots - 1.47240 \dots ∣ = 1.95413 \dots

Δ u_{11} = 1.95413 \dots

无法得出解

解为

u \approx 0.58141 \dots, u \approx - 1.23238 \dots

解为

u = - 1, u \approx 0.58141 \dots, u \approx - 1.23238 \dots

u = cos (x)

代回

cos (x) = - 1, cos (x) \approx 0.58141 \dots, cos (x) \approx - 1.23238 \dots

cos (x) = - 1, cos (x) \approx 0.58141 \dots, cos (x) \approx - 1.23238 \dots

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

cos (x) = - 1

的通解

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

x = π + 2 πn

x = π + 2 πn

cos (x) = 0.58141 \dots : x = arccos (0.58141 \dots) + 2 πn, x = 2 π - arccos (0.58141 \dots) + 2 πn

cos (x) = 0.58141 \dots

使用反三角函数性质

cos (x) = 0.58141 \dots

cos (x) = 0.58141 \dots

的通解

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

x = arccos (0.58141 \dots) + 2 πn, x = 2 π - arccos (0.58141 \dots) + 2 πn

x = arccos (0.58141 \dots) + 2 πn, x = 2 π - arccos (0.58141 \dots) + 2 πn

cos (x) = - 1.23238 \dots :

无解

cos (x) = - 1.23238 \dots

- 1 \leq cos (x) \leq 1

无解

合并所有解

x = π + 2 πn, x = arccos (0.58141 \dots) + 2 πn, x = 2 π - arccos (0.58141 \dots) + 2 πn

以小数形式表示解

x = π + 2 πn, x = 0.95033 \dots + 2 πn, x = 2 π - 0.95033 \dots + 2 πn

作图

流行的例子

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solvefor x,cot^2(x)= 1/3 求解

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

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tan^{5} (x) tan^{2} (x) = 1