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受欢迎的三角函数 >

tan^2(x)-4sin(x)+4=0

解答

tan^{2} (x) - 4 sin (x) + 4 = 0

解答

x \in R 无解

求解步骤

tan^{2} (x) - 4 sin (x) + 4 = 0

两边加上

4 sin (x)

tan^{2} (x) + 4 = 4 sin (x)

两边进行平方

(tan^{2} (x) + 4)^{2} = (4 sin (x))^{2}

两边减去

(4 sin (x))^{2}

(tan^{2} (x) + 4)^{2} - 16 sin^{2} (x) = 0

使用三角恒等式改写

(4 + tan^{2} (x))^{2} - 16 sin^{2} (x)

使用基本三角恒等式:

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

= (4 + (\frac{sin ( x )}{cos ( x )})^{2})^{2} - 16 sin^{2} (x)

使用指数法则:

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

= (4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )})^{2} - 16 sin^{2} (x)

(4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )})^{2} - 16 sin^{2} (x) = 0

分解

(4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )})^{2} - 16 sin^{2} (x) : (4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} + 4 sin (x)) (4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} - 4 sin (x))

(4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )})^{2} - 16 sin^{2} (x)

将

16 sin^{2} (x)

改写为

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

16 sin^{2} (x)

使用指数法则:

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

16 = (2^{2})^{2}

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

使用指数法则:

a^{m} b^{m} = (ab)^{m}

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

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

= (4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )})^{2} - (2^{2} sin (x))^{2}

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

(4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )})^{2} - (2^{2} sin (x))^{2} = ((4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )}) + 2^{2} sin (x)) ((4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )}) - 2^{2} sin (x))

= ((4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )}) + 2^{2} sin (x)) ((4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )}) - 2^{2} sin (x))

整理后得

= (\frac{sin ^{2} ( x )}{cos ^{2} ( x )} + 4 sin (x) + 4) (\frac{sin ^{2} ( x )}{cos ^{2} ( x )} - 4 sin (x) + 4)

(4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} + 4 sin (x)) (4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} - 4 sin (x)) = 0

分别求解每个部分

4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} + 4 sin (x) = 0 or 4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} - 4 sin (x) = 0

4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} + 4 sin (x) = 0 :

无解

4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} + 4 sin (x) = 0

化简

4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} + 4 sin (x) : \frac{4 cos ^{2} ( x ) + sin ^{2} ( x ) + 4 cos ^{2} ( x ) sin ( x )}{cos ^{2} ( x )}

4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} + 4 sin (x)

将项转换为分式:

4 = \frac{4 cos ^{2} ( x )}{cos ^{2} ( x )}, 4 sin (x) = \frac{4 sin ( x ) cos ^{2} ( x )}{cos ^{2} ( x )}

= \frac{4 cos ^{2} ( x )}{cos ^{2} ( x )} + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} + \frac{4 sin ( x ) cos ^{2} ( x )}{cos ^{2} ( x )}

因为分母相等，所以合并分式:

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

= \frac{4 cos ^{2} ( x ) + sin ^{2} ( x ) + 4 sin ( x ) cos ^{2} ( x )}{cos ^{2} ( x )}

\frac{4 cos ^{2} ( x ) + sin ^{2} ( x ) + 4 cos ^{2} ( x ) sin ( x )}{cos ^{2} ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

4 cos^{2} (x) + sin^{2} (x) + 4 cos^{2} (x) sin (x) = 0

使用三角恒等式改写

sin^{2} (x) + 4 cos^{2} (x) + 4 cos^{2} (x) sin (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)

= sin^{2} (x) + 4 (1 - sin^{2} (x)) + 4 (1 - sin^{2} (x)) sin (x)

化简

sin^{2} (x) + 4 (1 - sin^{2} (x)) + 4 (1 - sin^{2} (x)) sin (x) : - 3 sin^{2} (x) + 4 sin (x) - 4 sin^{3} (x) + 4

sin^{2} (x) + 4 (1 - sin^{2} (x)) + 4 (1 - sin^{2} (x)) sin (x)

= sin^{2} (x) + 4 (1 - sin^{2} (x)) + 4 sin (x) (1 - sin^{2} (x))

乘开

4 (1 - sin^{2} (x)) : 4 - 4 sin^{2} (x)

4 (1 - sin^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 4, b = 1, c = sin^{2} (x)

= 4 \cdot 1 - 4 sin^{2} (x)

数字相乘:

4 \cdot 1 = 4

= 4 - 4 sin^{2} (x)

= sin^{2} (x) + 4 - 4 sin^{2} (x) + 4 (1 - sin^{2} (x)) sin (x)

乘开

4 sin (x) (1 - sin^{2} (x)) : 4 sin (x) - 4 sin^{3} (x)

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

使用分配律:

a (b - c) = ab - a c

a = 4 sin (x), b = 1, c = sin^{2} (x)

= 4 sin (x) \cdot 1 - 4 sin (x) sin^{2} (x)

= 4 \cdot 1 \cdot sin (x) - 4 sin^{2} (x) sin (x)

化简

4 \cdot 1 \cdot sin (x) - 4 sin^{2} (x) sin (x) : 4 sin (x) - 4 sin^{3} (x)

4 \cdot 1 \cdot sin (x) - 4 sin^{2} (x) sin (x)

4 \cdot 1 \cdot sin (x) = 4 sin (x)

4 \cdot 1 \cdot sin (x)

数字相乘:

4 \cdot 1 = 4

= 4 sin (x)

4 sin^{2} (x) sin (x) = 4 sin^{3} (x)

4 sin^{2} (x) sin (x)

使用指数法则:

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

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

= 4 sin^{2 + 1} (x)

数字相加:

2 + 1 = 3

= 4 sin^{3} (x)

= 4 sin (x) - 4 sin^{3} (x)

= 4 sin (x) - 4 sin^{3} (x)

= sin^{2} (x) + 4 - 4 sin^{2} (x) + 4 sin (x) - 4 sin^{3} (x)

化简

sin^{2} (x) + 4 - 4 sin^{2} (x) + 4 sin (x) - 4 sin^{3} (x) : - 3 sin^{2} (x) + 4 sin (x) - 4 sin^{3} (x) + 4

sin^{2} (x) + 4 - 4 sin^{2} (x) + 4 sin (x) - 4 sin^{3} (x)

对同类项分组

= sin^{2} (x) - 4 sin^{2} (x) + 4 sin (x) - 4 sin^{3} (x) + 4

同类项相加:

sin^{2} (x) - 4 sin^{2} (x) = - 3 sin^{2} (x)

= - 3 sin^{2} (x) + 4 sin (x) - 4 sin^{3} (x) + 4

= - 3 sin^{2} (x) + 4 sin (x) - 4 sin^{3} (x) + 4

= - 3 sin^{2} (x) + 4 sin (x) - 4 sin^{3} (x) + 4

4 - 3 sin^{2} (x) + 4 sin (x) - 4 sin^{3} (x) = 0

用替代法求解

4 - 3 sin^{2} (x) + 4 sin (x) - 4 sin^{3} (x) = 0

令

: sin (x) = u

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

4 - 3 u^{2} + 4 u - 4 u^{3} = 0 : u \approx 1.06659 \dots

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

改写成标准形式

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

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

使用牛顿-拉弗森方法找到

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

的一个解:

u \approx 1.06659 \dots

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

牛顿-拉弗森近似法定义

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

找到

f^{^{'}} (u) : - 12 u^{2} - 6 u + 4

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

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

\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^{2}) = 6 u

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

将常数提出:

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

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

使用幂法则:

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

= 3 \cdot 2 u^{2 - 1}

化简

= 6 u

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

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

将常数提出:

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

= 4 \frac{d u}{d u}

使用常见微分定则:

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

= 4 \cdot 1

化简

= 4

\frac{d}{d u} (4) = 0

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

常数微分:

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

= 0

= - 12 u^{2} - 6 u + 4 + 0

化简

= - 12 u^{2} - 6 u + 4

令

u_{0} = 1

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 1.07142 \dots : Δ u_{1} = 0.07142 \dots

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

f^{^{'}} (u_{0}) = - 12 \cdot 1^{2} - 6 \cdot 1 + 4 = - 14

u_{1} = 1.07142 \dots

Δ u_{1} = ∣ 1.07142 \dots - 1 ∣ = 0.07142 \dots

Δ u_{1} = 0.07142 \dots

u_{2} = 1.06661 \dots : Δ u_{2} = 0.00481 \dots

f (u_{1}) = - 4 \cdot 1.07142 \dots^{3} - 3 \cdot 1.07142 \dots^{2} + 4 \cdot 1.07142 \dots + 4 = - 0.07798 \dots

f^{^{'}} (u_{1}) = - 12 \cdot 1.07142 \dots^{2} - 6 \cdot 1.07142 \dots + 4 = - 16.20408 \dots

u_{2} = 1.06661 \dots

Δ u_{2} = ∣ 1.06661 \dots - 1.07142 \dots ∣ = 0.00481 \dots

Δ u_{2} = 0.00481 \dots

u_{3} = 1.06659 \dots : Δ u_{3} = 0.00002 \dots

f (u_{2}) = - 4 \cdot 1.06661 \dots^{3} - 3 \cdot 1.06661 \dots^{2} + 4 \cdot 1.06661 \dots + 4 = - 0.00036 \dots

f^{^{'}} (u_{2}) = - 12 \cdot 1.06661 \dots^{2} - 6 \cdot 1.06661 \dots + 4 = - 16.05172 \dots

u_{3} = 1.06659 \dots

Δ u_{3} = ∣ 1.06659 \dots - 1.06661 \dots ∣ = 0.00002 \dots

Δ u_{3} = 0.00002 \dots

u_{4} = 1.06659 \dots : Δ u_{4} = 5.14172 E - 10

f (u_{3}) = - 4 \cdot 1.06659 \dots^{3} - 3 \cdot 1.06659 \dots^{2} + 4 \cdot 1.06659 \dots + 4 = - 8.25297 E - 9

f^{^{'}} (u_{3}) = - 12 \cdot 1.06659 \dots^{2} - 6 \cdot 1.06659 \dots + 4 = - 16.05100 \dots

u_{4} = 1.06659 \dots

Δ u_{4} = ∣ 1.06659 \dots - 1.06659 \dots ∣ = 5.14172 E - 10

Δ u_{4} = 5.14172 E - 10

u \approx 1.06659 \dots

使用长除法

Eq u a t i o n 0 : \frac{- 4 u ^{3} - 3 u ^{2} + 4 u + 4}{u - 1.06659 \dots} = - 4 u^{2} - 7.26637 \dots u - 3.75025 \dots

- 4 u^{2} - 7.26637 \dots u - 3.75025 \dots \approx 0

使用牛顿-拉弗森方法找到

- 4 u^{2} - 7.26637 \dots u - 3.75025 \dots = 0

的一个解:

u \in R

无解

- 4 u^{2} - 7.26637 \dots u - 3.75025 \dots = 0

牛顿-拉弗森近似法定义

f (u) = - 4 u^{2} - 7.26637 \dots u - 3.75025 \dots

找到

f^{^{'}} (u) : - 8 u - 7.26637 \dots

\frac{d}{d u} (- 4 u^{2} - 7.26637 \dots u - 3.75025 \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} (7.26637 \dots u) - \frac{d}{d u} (3.75025 \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} (7.26637 \dots u) = 7.26637 \dots

\frac{d}{d u} (7.26637 \dots u)

将常数提出:

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

= 7.26637 \dots \frac{d u}{d u}

使用常见微分定则:

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

= 7.26637 \dots \cdot 1

化简

= 7.26637 \dots

\frac{d}{d u} (3.75025 \dots) = 0

\frac{d}{d u} (3.75025 \dots)

常数微分:

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

= 0

= - 8 u - 7.26637 \dots - 0

化简

= - 8 u - 7.26637 \dots

令

u_{0} = - 1

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = - 0.34041 \dots : Δ u_{1} = 0.65958 \dots

f (u_{0}) = - 4 (- 1)^{2} - 7.26637 \dots (- 1) - 3.75025 \dots = - 0.48388 \dots

f^{^{'}} (u_{0}) = - 8 (- 1) - 7.26637 \dots = 0.73362 \dots

u_{1} = - 0.34041 \dots

Δ u_{1} = ∣ - 0.34041 \dots - (- 1) ∣ = 0.65958 \dots

Δ u_{1} = 0.65958 \dots

u_{2} = - 0.72346 \dots : Δ u_{2} = 0.38304 \dots

f (u_{1}) = - 4 (- 0.34041 \dots)^{2} - 7.26637 \dots (- 0.34041 \dots) - 3.75025 \dots = - 1.74019 \dots

f^{^{'}} (u_{1}) = - 8 (- 0.34041 \dots) - 7.26637 \dots = - 4.54302 \dots

u_{2} = - 0.72346 \dots

Δ u_{2} = ∣ - 0.72346 \dots - (- 0.34041 \dots) ∣ = 0.38304 \dots

Δ u_{2} = 0.38304 \dots

u_{3} = - 1.12038 \dots : Δ u_{3} = 0.39691 \dots

f (u_{2}) = - 4 (- 0.72346 \dots)^{2} - 7.26637 \dots (- 0.72346 \dots) - 3.75025 \dots = - 0.58690 \dots

f^{^{'}} (u_{2}) = - 8 (- 0.72346 \dots) - 7.26637 \dots = - 1.47865 \dots

u_{3} = - 1.12038 \dots

Δ u_{3} = ∣ - 1.12038 \dots - (- 0.72346 \dots) ∣ = 0.39691 \dots

Δ u_{3} = 0.39691 \dots

u_{4} = - 0.74896 \dots : Δ u_{4} = 0.37141 \dots

f (u_{3}) = - 4 (- 1.12038 \dots)^{2} - 7.26637 \dots (- 1.12038 \dots) - 3.75025 \dots = - 0.63017 \dots

f^{^{'}} (u_{3}) = - 8 (- 1.12038 \dots) - 7.26637 \dots = 1.69668 \dots

u_{4} = - 0.74896 \dots

Δ u_{4} = ∣ - 0.74896 \dots - (- 1.12038 \dots) ∣ = 0.37141 \dots

Δ u_{4} = 0.37141 \dots

u_{5} = - 1.18187 \dots : Δ u_{5} = 0.43290 \dots

f (u_{4}) = - 4 (- 0.74896 \dots)^{2} - 7.26637 \dots (- 0.74896 \dots) - 3.75025 \dots = - 0.55179 \dots

f^{^{'}} (u_{4}) = - 8 (- 0.74896 \dots) - 7.26637 \dots = - 1.27462 \dots

u_{5} = - 1.18187 \dots

Δ u_{5} = ∣ - 1.18187 \dots - (- 0.74896 \dots) ∣ = 0.43290 \dots

Δ u_{5} = 0.43290 \dots

u_{6} = - 0.83936 \dots : Δ u_{6} = 0.34251 \dots

f (u_{5}) = - 4 (- 1.18187 \dots)^{2} - 7.26637 \dots (- 1.18187 \dots) - 3.75025 \dots = - 0.74962 \dots

f^{^{'}} (u_{5}) = - 8 (- 1.18187 \dots) - 7.26637 \dots = 2.18860 \dots

u_{6} = - 0.83936 \dots

Δ u_{6} = ∣ - 0.83936 \dots - (- 1.18187 \dots) ∣ = 0.34251 \dots

Δ u_{6} = 0.34251 \dots

u_{7} = - 1.69025 \dots : Δ u_{7} = 0.85089 \dots

f (u_{6}) = - 4 (- 0.83936 \dots)^{2} - 7.26637 \dots (- 0.83936 \dots) - 3.75025 \dots = - 0.46925 \dots

f^{^{'}} (u_{6}) = - 8 (- 0.83936 \dots) - 7.26637 \dots = - 0.55149 \dots

u_{7} = - 1.69025 \dots

Δ u_{7} = ∣ - 1.69025 \dots - (- 0.83936 \dots) ∣ = 0.85089 \dots

Δ u_{7} = 0.85089 \dots

u_{8} = - 1.22729 \dots : Δ u_{8} = 0.46295 \dots

f (u_{7}) = - 4 (- 1.69025 \dots)^{2} - 7.26637 \dots (- 1.69025 \dots) - 3.75025 \dots = - 2.89606 \dots

f^{^{'}} (u_{7}) = - 8 (- 1.69025 \dots) - 7.26637 \dots = 6.25564 \dots

u_{8} = - 1.22729 \dots

Δ u_{8} = ∣ - 1.22729 \dots - (- 1.69025 \dots) ∣ = 0.46295 \dots

Δ u_{8} = 0.46295 \dots

u_{9} = - 0.89136 \dots : Δ u_{9} = 0.33593 \dots

f (u_{8}) = - 4 (- 1.22729 \dots)^{2} - 7.26637 \dots (- 1.22729 \dots) - 3.75025 \dots = - 0.85730 \dots

f^{^{'}} (u_{8}) = - 8 (- 1.22729 \dots) - 7.26637 \dots = 2.55202 \dots

u_{9} = - 0.89136 \dots

Δ u_{9} = ∣ - 0.89136 \dots - (- 1.22729 \dots) ∣ = 0.33593 \dots

Δ u_{9} = 0.33593 \dots

u_{10} = - 4.22467 \dots : Δ u_{10} = 3.33330 \dots

f (u_{9}) = - 4 (- 0.89136 \dots)^{2} - 7.26637 \dots (- 0.89136 \dots) - 3.75025 \dots = - 0.45139 \dots

f^{^{'}} (u_{9}) = - 8 (- 0.89136 \dots) - 7.26637 \dots = - 0.13541 \dots

u_{10} = - 4.22467 \dots

Δ u_{10} = ∣ - 4.22467 \dots - (- 0.89136 \dots) ∣ = 3.33330 \dots

Δ u_{10} = 3.33330 \dots

无法得出解

解是

u \approx 1.06659 \dots

u = sin (x)

代回

sin (x) \approx 1.06659 \dots

sin (x) \approx 1.06659 \dots

sin (x) = 1.06659 \dots :

无解

sin (x) = 1.06659 \dots

- 1 \leq sin (x) \leq 1

无解

合并所有解

无解

4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} - 4 sin (x) = 0 :

无解

4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} - 4 sin (x) = 0

化简

4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} - 4 sin (x) : \frac{4 cos ^{2} ( x ) + sin ^{2} ( x ) - 4 cos ^{2} ( x ) sin ( x )}{cos ^{2} ( x )}

4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} - 4 sin (x)

将项转换为分式:

4 = \frac{4 cos ^{2} ( x )}{cos ^{2} ( x )}, 4 sin (x) = \frac{4 sin ( x ) cos ^{2} ( x )}{cos ^{2} ( x )}

= \frac{4 cos ^{2} ( x )}{cos ^{2} ( x )} + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} - \frac{4 sin ( x ) cos ^{2} ( x )}{cos ^{2} ( x )}

因为分母相等，所以合并分式:

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

= \frac{4 cos ^{2} ( x ) + sin ^{2} ( x ) - 4 sin ( x ) cos ^{2} ( x )}{cos ^{2} ( x )}

\frac{4 cos ^{2} ( x ) + sin ^{2} ( x ) - 4 cos ^{2} ( x ) sin ( x )}{cos ^{2} ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

4 cos^{2} (x) + sin^{2} (x) - 4 cos^{2} (x) sin (x) = 0

使用三角恒等式改写

sin^{2} (x) + 4 cos^{2} (x) - 4 cos^{2} (x) sin (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)

= sin^{2} (x) + 4 (1 - sin^{2} (x)) - 4 (1 - sin^{2} (x)) sin (x)

化简

sin^{2} (x) + 4 (1 - sin^{2} (x)) - 4 (1 - sin^{2} (x)) sin (x) : - 3 sin^{2} (x) - 4 sin (x) + 4 sin^{3} (x) + 4

sin^{2} (x) + 4 (1 - sin^{2} (x)) - 4 (1 - sin^{2} (x)) sin (x)

= sin^{2} (x) + 4 (1 - sin^{2} (x)) - 4 sin (x) (1 - sin^{2} (x))

乘开

4 (1 - sin^{2} (x)) : 4 - 4 sin^{2} (x)

4 (1 - sin^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 4, b = 1, c = sin^{2} (x)

= 4 \cdot 1 - 4 sin^{2} (x)

数字相乘:

4 \cdot 1 = 4

= 4 - 4 sin^{2} (x)

= sin^{2} (x) + 4 - 4 sin^{2} (x) - 4 (1 - sin^{2} (x)) sin (x)

乘开

- 4 sin (x) (1 - sin^{2} (x)) : - 4 sin (x) + 4 sin^{3} (x)

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

使用分配律:

a (b - c) = ab - a c

a = - 4 sin (x), b = 1, c = sin^{2} (x)

= - 4 sin (x) \cdot 1 - (- 4 sin (x)) sin^{2} (x)

使用加减运算法则

- (- a) = a

= - 4 \cdot 1 \cdot sin (x) + 4 sin^{2} (x) sin (x)

化简

- 4 \cdot 1 \cdot sin (x) + 4 sin^{2} (x) sin (x) : - 4 sin (x) + 4 sin^{3} (x)

- 4 \cdot 1 \cdot sin (x) + 4 sin^{2} (x) sin (x)

4 \cdot 1 \cdot sin (x) = 4 sin (x)

4 \cdot 1 \cdot sin (x)

数字相乘:

4 \cdot 1 = 4

= 4 sin (x)

4 sin^{2} (x) sin (x) = 4 sin^{3} (x)

4 sin^{2} (x) sin (x)

使用指数法则:

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

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

= 4 sin^{2 + 1} (x)

数字相加:

2 + 1 = 3

= 4 sin^{3} (x)

= - 4 sin (x) + 4 sin^{3} (x)

= - 4 sin (x) + 4 sin^{3} (x)

= sin^{2} (x) + 4 - 4 sin^{2} (x) - 4 sin (x) + 4 sin^{3} (x)

化简

sin^{2} (x) + 4 - 4 sin^{2} (x) - 4 sin (x) + 4 sin^{3} (x) : - 3 sin^{2} (x) - 4 sin (x) + 4 sin^{3} (x) + 4

sin^{2} (x) + 4 - 4 sin^{2} (x) - 4 sin (x) + 4 sin^{3} (x)

对同类项分组

= sin^{2} (x) - 4 sin^{2} (x) - 4 sin (x) + 4 sin^{3} (x) + 4

同类项相加:

sin^{2} (x) - 4 sin^{2} (x) = - 3 sin^{2} (x)

= - 3 sin^{2} (x) - 4 sin (x) + 4 sin^{3} (x) + 4

= - 3 sin^{2} (x) - 4 sin (x) + 4 sin^{3} (x) + 4

= - 3 sin^{2} (x) - 4 sin (x) + 4 sin^{3} (x) + 4

4 - 3 sin^{2} (x) - 4 sin (x) + 4 sin^{3} (x) = 0

用替代法求解

4 - 3 sin^{2} (x) - 4 sin (x) + 4 sin^{3} (x) = 0

令

: sin (x) = u

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

4 - 3 u^{2} - 4 u + 4 u^{3} = 0 : u \approx - 1.06659 \dots

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

改写成标准形式

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

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

使用牛顿-拉弗森方法找到

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

的一个解:

u \approx - 1.06659 \dots

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

牛顿-拉弗森近似法定义

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

找到

f^{^{'}} (u) : 12 u^{2} - 6 u - 4

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

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

\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^{2}) = 6 u

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

将常数提出:

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

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

使用幂法则:

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

= 3 \cdot 2 u^{2 - 1}

化简

= 6 u

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

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

将常数提出:

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

= 4 \frac{d u}{d u}

使用常见微分定则:

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

= 4 \cdot 1

化简

= 4

\frac{d}{d u} (4) = 0

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

常数微分:

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

= 0

= 12 u^{2} - 6 u - 4 + 0

化简

= 12 u^{2} - 6 u - 4

令

u_{0} = - 1

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = - 1.07142 \dots : Δ u_{1} = 0.07142 \dots

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

f^{^{'}} (u_{0}) = 12 (- 1)^{2} - 6 (- 1) - 4 = 14

u_{1} = - 1.07142 \dots

Δ u_{1} = ∣ - 1.07142 \dots - (- 1) ∣ = 0.07142 \dots

Δ u_{1} = 0.07142 \dots

u_{2} = - 1.06661 \dots : Δ u_{2} = 0.00481 \dots

f (u_{1}) = 4 (- 1.07142 \dots)^{3} - 3 (- 1.07142 \dots)^{2} - 4 (- 1.07142 \dots) + 4 = - 0.07798 \dots

f^{^{'}} (u_{1}) = 12 (- 1.07142 \dots)^{2} - 6 (- 1.07142 \dots) - 4 = 16.20408 \dots

u_{2} = - 1.06661 \dots

Δ u_{2} = ∣ - 1.06661 \dots - (- 1.07142 \dots) ∣ = 0.00481 \dots

Δ u_{2} = 0.00481 \dots

u_{3} = - 1.06659 \dots : Δ u_{3} = 0.00002 \dots

f (u_{2}) = 4 (- 1.06661 \dots)^{3} - 3 (- 1.06661 \dots)^{2} - 4 (- 1.06661 \dots) + 4 = - 0.00036 \dots

f^{^{'}} (u_{2}) = 12 (- 1.06661 \dots)^{2} - 6 (- 1.06661 \dots) - 4 = 16.05172 \dots

u_{3} = - 1.06659 \dots

Δ u_{3} = ∣ - 1.06659 \dots - (- 1.06661 \dots) ∣ = 0.00002 \dots

Δ u_{3} = 0.00002 \dots

u_{4} = - 1.06659 \dots : Δ u_{4} = 5.14172 E - 10

f (u_{3}) = 4 (- 1.06659 \dots)^{3} - 3 (- 1.06659 \dots)^{2} - 4 (- 1.06659 \dots) + 4 = - 8.25297 E - 9

f^{^{'}} (u_{3}) = 12 (- 1.06659 \dots)^{2} - 6 (- 1.06659 \dots) - 4 = 16.05100 \dots

u_{4} = - 1.06659 \dots

Δ u_{4} = ∣ - 1.06659 \dots - (- 1.06659 \dots) ∣ = 5.14172 E - 10

Δ u_{4} = 5.14172 E - 10

u \approx - 1.06659 \dots

使用长除法

Eq u a t i o n 0 : \frac{4 u ^{3} - 3 u ^{2} - 4 u + 4}{u + 1.06659 \dots} = 4 u^{2} - 7.26637 \dots u + 3.75025 \dots

4 u^{2} - 7.26637 \dots u + 3.75025 \dots \approx 0

使用牛顿-拉弗森方法找到

4 u^{2} - 7.26637 \dots u + 3.75025 \dots = 0

的一个解:

u \in R

无解

4 u^{2} - 7.26637 \dots u + 3.75025 \dots = 0

牛顿-拉弗森近似法定义

f (u) = 4 u^{2} - 7.26637 \dots u + 3.75025 \dots

找到

f^{^{'}} (u) : 8 u - 7.26637 \dots

\frac{d}{d u} (4 u^{2} - 7.26637 \dots u + 3.75025 \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} (7.26637 \dots u) + \frac{d}{d u} (3.75025 \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} (7.26637 \dots u) = 7.26637 \dots

\frac{d}{d u} (7.26637 \dots u)

将常数提出:

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

= 7.26637 \dots \frac{d u}{d u}

使用常见微分定则:

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

= 7.26637 \dots \cdot 1

化简

= 7.26637 \dots

\frac{d}{d u} (3.75025 \dots) = 0

\frac{d}{d u} (3.75025 \dots)

常数微分:

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

= 0

= 8 u - 7.26637 \dots + 0

化简

= 8 u - 7.26637 \dots

令

u_{0} = 1

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 0.34041 \dots : Δ u_{1} = 0.65958 \dots

f (u_{0}) = 4 \cdot 1^{2} - 7.26637 \dots \cdot 1 + 3.75025 \dots = 0.48388 \dots

f^{^{'}} (u_{0}) = 8 \cdot 1 - 7.26637 \dots = 0.73362 \dots

u_{1} = 0.34041 \dots

Δ u_{1} = ∣ 0.34041 \dots - 1 ∣ = 0.65958 \dots

Δ u_{1} = 0.65958 \dots

u_{2} = 0.72346 \dots : Δ u_{2} = 0.38304 \dots

f (u_{1}) = 4 \cdot 0.34041 \dots^{2} - 7.26637 \dots \cdot 0.34041 \dots + 3.75025 \dots = 1.74019 \dots

f^{^{'}} (u_{1}) = 8 \cdot 0.34041 \dots - 7.26637 \dots = - 4.54302 \dots

u_{2} = 0.72346 \dots

Δ u_{2} = ∣ 0.72346 \dots - 0.34041 \dots ∣ = 0.38304 \dots

Δ u_{2} = 0.38304 \dots

u_{3} = 1.12038 \dots : Δ u_{3} = 0.39691 \dots

f (u_{2}) = 4 \cdot 0.72346 \dots^{2} - 7.26637 \dots \cdot 0.72346 \dots + 3.75025 \dots = 0.58690 \dots

f^{^{'}} (u_{2}) = 8 \cdot 0.72346 \dots - 7.26637 \dots = - 1.47865 \dots

u_{3} = 1.12038 \dots

Δ u_{3} = ∣ 1.12038 \dots - 0.72346 \dots ∣ = 0.39691 \dots

Δ u_{3} = 0.39691 \dots

u_{4} = 0.74896 \dots : Δ u_{4} = 0.37141 \dots

f (u_{3}) = 4 \cdot 1.12038 \dots^{2} - 7.26637 \dots \cdot 1.12038 \dots + 3.75025 \dots = 0.63017 \dots

f^{^{'}} (u_{3}) = 8 \cdot 1.12038 \dots - 7.26637 \dots = 1.69668 \dots

u_{4} = 0.74896 \dots

Δ u_{4} = ∣ 0.74896 \dots - 1.12038 \dots ∣ = 0.37141 \dots

Δ u_{4} = 0.37141 \dots

u_{5} = 1.18187 \dots : Δ u_{5} = 0.43290 \dots

f (u_{4}) = 4 \cdot 0.74896 \dots^{2} - 7.26637 \dots \cdot 0.74896 \dots + 3.75025 \dots = 0.55179 \dots

f^{^{'}} (u_{4}) = 8 \cdot 0.74896 \dots - 7.26637 \dots = - 1.27462 \dots

u_{5} = 1.18187 \dots

Δ u_{5} = ∣ 1.18187 \dots - 0.74896 \dots ∣ = 0.43290 \dots

Δ u_{5} = 0.43290 \dots

u_{6} = 0.83936 \dots : Δ u_{6} = 0.34251 \dots

f (u_{5}) = 4 \cdot 1.18187 \dots^{2} - 7.26637 \dots \cdot 1.18187 \dots + 3.75025 \dots = 0.74962 \dots

f^{^{'}} (u_{5}) = 8 \cdot 1.18187 \dots - 7.26637 \dots = 2.18860 \dots

u_{6} = 0.83936 \dots

Δ u_{6} = ∣ 0.83936 \dots - 1.18187 \dots ∣ = 0.34251 \dots

Δ u_{6} = 0.34251 \dots

u_{7} = 1.69025 \dots : Δ u_{7} = 0.85089 \dots

f (u_{6}) = 4 \cdot 0.83936 \dots^{2} - 7.26637 \dots \cdot 0.83936 \dots + 3.75025 \dots = 0.46925 \dots

f^{^{'}} (u_{6}) = 8 \cdot 0.83936 \dots - 7.26637 \dots = - 0.55149 \dots

u_{7} = 1.69025 \dots

Δ u_{7} = ∣ 1.69025 \dots - 0.83936 \dots ∣ = 0.85089 \dots

Δ u_{7} = 0.85089 \dots

u_{8} = 1.22729 \dots : Δ u_{8} = 0.46295 \dots

f (u_{7}) = 4 \cdot 1.69025 \dots^{2} - 7.26637 \dots \cdot 1.69025 \dots + 3.75025 \dots = 2.89606 \dots

f^{^{'}} (u_{7}) = 8 \cdot 1.69025 \dots - 7.26637 \dots = 6.25564 \dots

u_{8} = 1.22729 \dots

Δ u_{8} = ∣ 1.22729 \dots - 1.69025 \dots ∣ = 0.46295 \dots

Δ u_{8} = 0.46295 \dots

u_{9} = 0.89136 \dots : Δ u_{9} = 0.33593 \dots

f (u_{8}) = 4 \cdot 1.22729 \dots^{2} - 7.26637 \dots \cdot 1.22729 \dots + 3.75025 \dots = 0.85730 \dots

f^{^{'}} (u_{8}) = 8 \cdot 1.22729 \dots - 7.26637 \dots = 2.55202 \dots

u_{9} = 0.89136 \dots

Δ u_{9} = ∣ 0.89136 \dots - 1.22729 \dots ∣ = 0.33593 \dots

Δ u_{9} = 0.33593 \dots

u_{10} = 4.22467 \dots : Δ u_{10} = 3.33330 \dots

f (u_{9}) = 4 \cdot 0.89136 \dots^{2} - 7.26637 \dots \cdot 0.89136 \dots + 3.75025 \dots = 0.45139 \dots

f^{^{'}} (u_{9}) = 8 \cdot 0.89136 \dots - 7.26637 \dots = - 0.13541 \dots

u_{10} = 4.22467 \dots

Δ u_{10} = ∣ 4.22467 \dots - 0.89136 \dots ∣ = 3.33330 \dots

Δ u_{10} = 3.33330 \dots

无法得出解

解是

u \approx - 1.06659 \dots

u = sin (x)

代回

sin (x) \approx - 1.06659 \dots

sin (x) \approx - 1.06659 \dots

sin (x) = - 1.06659 \dots :

无解

sin (x) = - 1.06659 \dots

- 1 \leq sin (x) \leq 1

无解

合并所有解

无解

合并所有解

无解

将解代入原方程进行验证

将它们代入

tan^{2} (x) - 4 sin (x) + 4 = 0

检验解是否符合

去除与方程不符的解。

x \in R 无解

作图

流行的例子

1 - cos (2 x) = 0

sin (x) = \frac{37}{64}

15cos^2(x)+7cos(x)-2=0

15 cos^{2} (x) + 7 cos (x) - 2 = 0

2(sin(t))^2-sin(t)-1=0

2 (sin (t))^{2} - sin (t) - 1 = 0

cos (x) = 0.625