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受欢迎的 >

(sin(x))/(cos(x))=(sin(x))/(1+cot^2(x))

解答

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

解答

x \in R 无解

求解步骤

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

两边减去

\frac{sin ( x )}{1 + cot ^{2} ( x )}

\frac{sin ( x )}{cos ( x )} - \frac{sin ( x )}{1 + cot ^{2} ( x )} = 0

化简

\frac{sin ( x )}{cos ( x )} - \frac{sin ( x )}{1 + cot ^{2} ( x )} : \frac{sin ( x ) ( cot ^{2} ( x ) + 1 ) - sin ( x ) cos ( x )}{cos ( x ) ( cot ^{2} ( x ) + 1 )}

\frac{sin ( x )}{cos ( x )} - \frac{sin ( x )}{1 + cot ^{2} ( x )}

cos (x), 1 + cot^{2} (x)

的最小公倍数:

cos (x) (cot^{2} (x) + 1)

cos (x), 1 + cot^{2} (x)

最小公倍数 (LCM)

计算出由出现在

cos (x)

或

1 + cot^{2} (x)

中的因子组成的表达式

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

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

cos (x) (cot^{2} (x) + 1)

对于

\frac{sin ( x )}{cos ( x )} :

将分母和分子乘以

cot^{2} (x) + 1

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

对于

\frac{sin ( x )}{1 + cot ^{2} ( x )} :

将分母和分子乘以

cos (x)

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

= \frac{sin ( x ) ( cot ^{2} ( x ) + 1 )}{cos ( x ) ( cot ^{2} ( x ) + 1 )} - \frac{sin ( x ) cos ( x )}{( 1 + cot ^{2} ( x ) ) cos ( 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{sin ( x ) ( cot ^{2} ( x ) + 1 ) - sin ( x ) cos ( x )}{cos ( x ) ( cot ^{2} ( x ) + 1 )}

\frac{sin ( x ) ( cot ^{2} ( x ) + 1 ) - sin ( x ) cos ( x )}{cos ( x ) ( cot ^{2} ( x ) + 1 )} = 0

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

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

使用三角恒等式改写

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

使用基本三角恒等式:

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

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

使用指数法则:

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

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

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

分解

(1 + \frac{cos ^{2} ( x )}{sin ^{2} ( x )}) sin (x) - cos (x) sin (x) : sin (x) (1 + \frac{cos ^{2} ( x )}{sin ^{2} ( x )} - cos (x))

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

因式分解出通项

sin (x)

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

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

分别求解每个部分

sin (x) = 0 or 1 + \frac{cos ^{2} ( x )}{sin ^{2} ( x )} - cos (x) = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

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

无解

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

化简

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

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

将项转换为分式:

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

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

乘以:

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

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

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

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

化简

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

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

乘开

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

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

使用分配律:

a (b - c) = ab - a c

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

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

使用加减运算法则

- (- a) = a

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

化简

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

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

1 \cdot cos (x) = cos (x)

1 \cdot cos (x)

乘以:

1 \cdot cos (x) = cos (x)

= cos (x)

cos^{2} (x) cos (x) = cos^{3} (x)

cos^{2} (x) cos (x)

使用指数法则:

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

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

= cos^{2 + 1} (x)

数字相加:

2 + 1 = 3

= cos^{3} (x)

= - cos (x) + cos^{3} (x)

= - cos (x) + cos^{3} (x)

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

化简

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

cos^{2} (x) + 1 - cos^{2} (x) - cos (x) + cos^{3} (x)

对同类项分组

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

同类项相加:

cos^{2} (x) - cos^{2} (x) = 0

= - cos (x) + cos^{3} (x) + 1

= - cos (x) + cos^{3} (x) + 1

= - cos (x) + cos^{3} (x) + 1

1 - cos (x) + cos^{3} (x) = 0

用替代法求解

1 - cos (x) + cos^{3} (x) = 0

令

: cos (x) = u

1 - u + u^{3} = 0

1 - u + u^{3} = 0 : u \approx - 1.32471 \dots

1 - u + u^{3} = 0

改写成标准形式

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

u^{3} - u + 1 = 0

使用牛顿-拉弗森方法找到

u^{3} - u + 1 = 0

的一个解:

u \approx - 1.32471 \dots

u^{3} - u + 1 = 0

牛顿-拉弗森近似法定义

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

找到

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

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

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} (1)

\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} (1) = 0

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

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} = - 1.5 : Δ u_{1} = 0.5

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

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

u_{1} = - 1.5

Δ u_{1} = ∣ - 1.5 - (- 1) ∣ = 0.5

Δ u_{1} = 0.5

u_{2} = - 1.34782 \dots : Δ u_{2} = 0.15217 \dots

f (u_{1}) = (- 1.5)^{3} - (- 1.5) + 1 = - 0.875

f^{^{'}} (u_{1}) = 3 (- 1.5)^{2} - 1 = 5.75

u_{2} = - 1.34782 \dots

Δ u_{2} = ∣ - 1.34782 \dots - (- 1.5) ∣ = 0.15217 \dots

Δ u_{2} = 0.15217 \dots

u_{3} = - 1.32520 \dots : Δ u_{3} = 0.02262 \dots

f (u_{2}) = (- 1.34782 \dots)^{3} - (- 1.34782 \dots) + 1 = - 0.10068 \dots

f^{^{'}} (u_{2}) = 3 (- 1.34782 \dots)^{2} - 1 = 4.44990 \dots

u_{3} = - 1.32520 \dots

Δ u_{3} = ∣ - 1.32520 \dots - (- 1.34782 \dots) ∣ = 0.02262 \dots

Δ u_{3} = 0.02262 \dots

u_{4} = - 1.32471 \dots : Δ u_{4} = 0.00048 \dots

f (u_{3}) = (- 1.32520 \dots)^{3} - (- 1.32520 \dots) + 1 = - 0.00205 \dots

f^{^{'}} (u_{3}) = 3 (- 1.32520 \dots)^{2} - 1 = 4.26846 \dots

u_{4} = - 1.32471 \dots

Δ u_{4} = ∣ - 1.32471 \dots - (- 1.32520 \dots) ∣ = 0.00048 \dots

Δ u_{4} = 0.00048 \dots

u_{5} = - 1.32471 \dots : Δ u_{5} = 2.16754 E - 7

f (u_{4}) = (- 1.32471 \dots)^{3} - (- 1.32471 \dots) + 1 = - 9.24378 E - 7

f^{^{'}} (u_{4}) = 3 (- 1.32471 \dots)^{2} - 1 = 4.26463 \dots

u_{5} = - 1.32471 \dots

Δ u_{5} = ∣ - 1.32471 \dots - (- 1.32471 \dots) ∣ = 2.16754 E - 7

Δ u_{5} = 2.16754 E - 7

u \approx - 1.32471 \dots

使用长除法

Eq u a t i o n 0 : \frac{u ^{3} - u + 1}{u + 1.32471 \dots} = u^{2} - 1.32471 \dots u + 0.75487 \dots

u^{2} - 1.32471 \dots u + 0.75487 \dots \approx 0

使用牛顿-拉弗森方法找到

u^{2} - 1.32471 \dots u + 0.75487 \dots = 0

的一个解:

u \in R

无解

u^{2} - 1.32471 \dots u + 0.75487 \dots = 0

牛顿-拉弗森近似法定义

f (u) = u^{2} - 1.32471 \dots u + 0.75487 \dots

找到

f^{^{'}} (u) : 2 u - 1.32471 \dots

\frac{d}{d u} (u^{2} - 1.32471 \dots u + 0.75487 \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} (1.32471 \dots u) + \frac{d}{d u} (0.75487 \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} (1.32471 \dots u) = 1.32471 \dots

\frac{d}{d u} (1.32471 \dots u)

将常数提出:

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

= 1.32471 \dots \frac{d u}{d u}

使用常见微分定则:

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

= 1.32471 \dots \cdot 1

化简

= 1.32471 \dots

\frac{d}{d u} (0.75487 \dots) = 0

\frac{d}{d u} (0.75487 \dots)

常数微分:

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

= 0

= 2 u - 1.32471 \dots + 0

化简

= 2 u - 1.32471 \dots

令

u_{0} = 1

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 0.36299 \dots : Δ u_{1} = 0.63700 \dots

f (u_{0}) = 1^{2} - 1.32471 \dots \cdot 1 + 0.75487 \dots = 0.43015 \dots

f^{^{'}} (u_{0}) = 2 \cdot 1 - 1.32471 \dots = 0.67528 \dots

u_{1} = 0.36299 \dots

Δ u_{1} = ∣ 0.36299 \dots - 1 ∣ = 0.63700 \dots

Δ u_{1} = 0.63700 \dots

u_{2} = 1.04072 \dots : Δ u_{2} = 0.67772 \dots

f (u_{1}) = 0.36299 \dots^{2} - 1.32471 \dots \cdot 0.36299 \dots + 0.75487 \dots = 0.40577 \dots

f^{^{'}} (u_{1}) = 2 \cdot 0.36299 \dots - 1.32471 \dots = - 0.59873 \dots

u_{2} = 1.04072 \dots

Δ u_{2} = ∣ 1.04072 \dots - 0.36299 \dots ∣ = 0.67772 \dots

Δ u_{2} = 0.67772 \dots

u_{3} = 0.43374 \dots : Δ u_{3} = 0.60697 \dots

f (u_{2}) = 1.04072 \dots^{2} - 1.32471 \dots \cdot 1.04072 \dots + 0.75487 \dots = 0.45931 \dots

f^{^{'}} (u_{2}) = 2 \cdot 1.04072 \dots - 1.32471 \dots = 0.75672 \dots

u_{3} = 0.43374 \dots

Δ u_{3} = ∣ 0.43374 \dots - 1.04072 \dots ∣ = 0.60697 \dots

Δ u_{3} = 0.60697 \dots

u_{4} = 1.23950 \dots : Δ u_{4} = 0.80576 \dots

f (u_{3}) = 0.43374 \dots^{2} - 1.32471 \dots \cdot 0.43374 \dots + 0.75487 \dots = 0.36842 \dots

f^{^{'}} (u_{3}) = 2 \cdot 0.43374 \dots - 1.32471 \dots = - 0.45723 \dots

u_{4} = 1.23950 \dots

Δ u_{4} = ∣ 1.23950 \dots - 0.43374 \dots ∣ = 0.80576 \dots

Δ u_{4} = 0.80576 \dots

u_{5} = 0.67703 \dots : Δ u_{5} = 0.56247 \dots

f (u_{4}) = 1.23950 \dots^{2} - 1.32471 \dots \cdot 1.23950 \dots + 0.75487 \dots = 0.64925 \dots

f^{^{'}} (u_{4}) = 2 \cdot 1.23950 \dots - 1.32471 \dots = 1.15429 \dots

u_{5} = 0.67703 \dots

Δ u_{5} = ∣ 0.67703 \dots - 1.23950 \dots ∣ = 0.56247 \dots

Δ u_{5} = 0.56247 \dots

u_{6} = - 10.09982 \dots : Δ u_{6} = 10.77686 \dots

f (u_{5}) = 0.67703 \dots^{2} - 1.32471 \dots \cdot 0.67703 \dots + 0.75487 \dots = 0.31637 \dots

f^{^{'}} (u_{5}) = 2 \cdot 0.67703 \dots - 1.32471 \dots = 0.02935 \dots

u_{6} = - 10.09982 \dots

Δ u_{6} = ∣ - 10.09982 \dots - 0.67703 \dots ∣ = 10.77686 \dots

Δ u_{6} = 10.77686 \dots

u_{7} = - 4.70404 \dots : Δ u_{7} = 5.39578 \dots

f (u_{6}) = (- 10.09982 \dots)^{2} - 1.32471 \dots (- 10.09982 \dots) + 0.75487 \dots = 116.14080 \dots

f^{^{'}} (u_{6}) = 2 (- 10.09982 \dots) - 1.32471 \dots = - 21.52437 \dots

u_{7} = - 4.70404 \dots

Δ u_{7} = ∣ - 4.70404 \dots - (- 10.09982 \dots) ∣ = 5.39578 \dots

Δ u_{7} = 5.39578 \dots

u_{8} = - 1.99138 \dots : Δ u_{8} = 2.71265 \dots

f (u_{7}) = (- 4.70404 \dots)^{2} - 1.32471 \dots (- 4.70404 \dots) + 0.75487 \dots = 29.11445 \dots

f^{^{'}} (u_{7}) = 2 (- 4.70404 \dots) - 1.32471 \dots = - 10.73280 \dots

u_{8} = - 1.99138 \dots

Δ u_{8} = ∣ - 1.99138 \dots - (- 4.70404 \dots) ∣ = 2.71265 \dots

Δ u_{8} = 2.71265 \dots

u_{9} = - 0.60494 \dots : Δ u_{9} = 1.38644 \dots

f (u_{8}) = (- 1.99138 \dots)^{2} - 1.32471 \dots (- 1.99138 \dots) + 0.75487 \dots = 7.35852 \dots

f^{^{'}} (u_{8}) = 2 (- 1.99138 \dots) - 1.32471 \dots = - 5.30749 \dots

u_{9} = - 0.60494 \dots

Δ u_{9} = ∣ - 0.60494 \dots - (- 1.99138 \dots) ∣ = 1.38644 \dots

Δ u_{9} = 1.38644 \dots

u_{10} = 0.15344 \dots : Δ u_{10} = 0.75838 \dots

f (u_{9}) = (- 0.60494 \dots)^{2} - 1.32471 \dots (- 0.60494 \dots) + 0.75487 \dots = 1.92221 \dots

f^{^{'}} (u_{9}) = 2 (- 0.60494 \dots) - 1.32471 \dots = - 2.53460 \dots

u_{10} = 0.15344 \dots

Δ u_{10} = ∣ 0.15344 \dots - (- 0.60494 \dots) ∣ = 0.75838 \dots

Δ u_{10} = 0.75838 \dots

u_{11} = 0.71852 \dots : Δ u_{11} = 0.56507 \dots

f (u_{10}) = 0.15344 \dots^{2} - 1.32471 \dots \cdot 0.15344 \dots + 0.75487 \dots = 0.57515 \dots

f^{^{'}} (u_{10}) = 2 \cdot 0.15344 \dots - 1.32471 \dots = - 1.01783 \dots

u_{11} = 0.71852 \dots

Δ u_{11} = ∣ 0.71852 \dots - 0.15344 \dots ∣ = 0.56507 \dots

Δ u_{11} = 0.56507 \dots

u_{12} = - 2.12427 \dots : Δ u_{12} = 2.84279 \dots

f (u_{11}) = 0.71852 \dots^{2} - 1.32471 \dots \cdot 0.71852 \dots + 0.75487 \dots = 0.31931 \dots

f^{^{'}} (u_{11}) = 2 \cdot 0.71852 \dots - 1.32471 \dots = 0.11232 \dots

u_{12} = - 2.12427 \dots

Δ u_{12} = ∣ - 2.12427 \dots - 0.71852 \dots ∣ = 2.84279 \dots

Δ u_{12} = 2.84279 \dots

无法得出解

解是

u \approx - 1.32471 \dots

u = cos (x)

代回

cos (x) \approx - 1.32471 \dots

cos (x) \approx - 1.32471 \dots

cos (x) = - 1.32471 \dots :

无解

cos (x) = - 1.32471 \dots

- 1 \leq cos (x) \leq 1

无解

合并所有解

无解

合并所有解

x = 2 πn, x = π + 2 πn

因为方程对以下值无定义:

2 πn, π + 2 πn

x \in R 无解

流行的例子

3 + 8 cos (2 θ) = - 1

-4cos^2(x)+6cos(x)=2

- 4 cos^{2} (x) + 6 cos (x) = 2

sec (θ) = - \frac{4}{3}

tan(2x)=tan(2018)

tan (2 x) = tan (201 8^{\circ})

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

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