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受欢迎的三角函数 >

3tan(x)+cot(x)<5sin(x)

解答

3 tan (x) + cot (x) < 5 sin (x)

解答

\frac{π}{2} + 2 πn < x < π + 2 πn or \frac{3 π}{2} + 2 πn < x < 2 π + 2 πn

+2

间隔符号

(\frac{π}{2} + 2 πn, π + 2 πn) \cup (\frac{3 π}{2} + 2 πn, 2 π + 2 πn)

十进制

1.57079 \dots + 2 πn < x < 3.14159 \dots + 2 πn or 4.71238 \dots + 2 πn < x < 6.28318 \dots + 2 πn

求解步骤

3 tan (x) + cot (x) < 5 sin (x)

将

5 sin (x)

para o lado esquerdo

3 tan (x) + cot (x) < 5 sin (x)

两边减去

5 sin (x)

3 tan (x) + cot (x) - 5 sin (x) < 5 sin (x) - 5 sin (x)

3 tan (x) + cot (x) - 5 sin (x) < 0

3 tan (x) + cot (x) - 5 sin (x) < 0

3 tan (x) + cot (x) - 5 sin (x)

的周期:

2 π

周期函数和的复合周期是这些周期的最小公倍数

3 tan (x), cot (x), 5 sin (x)

3 tan (x)

的周期:

π

周期

t an (b x + c) + d = \frac{tan ( x ) 的周期}{∣ b ∣}

tan (x)

的周期是

π

= \frac{π}{∣ 1 ∣}

化简

= π

cot (x)

的周期:

π

cot (x)

的周期是

π

= π

5 sin (x)

的周期:

2 π

周期

s in (b x + c) + d = \frac{sin ( x ) 的周期}{∣ b ∣}

sin (x)

的周期是

2 π

= \frac{2 π}{∣ 1 ∣}

化简

= 2 π

合并周期:

π, π, 2 π

= 2 π

用 sin, cos 表示

3 tan (x) + cot (x) - 5 sin (x) < 0

使用基本三角恒等式:

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

3 \cdot \frac{sin ( x )}{cos ( x )} + cot (x) - 5 sin (x) < 0

使用基本三角恒等式:

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

3 \cdot \frac{sin ( x )}{cos ( x )} + \frac{cos ( x )}{sin ( x )} - 5 sin (x) < 0

3 \cdot \frac{sin ( x )}{cos ( x )} + \frac{cos ( x )}{sin ( x )} - 5 sin (x) < 0

化简

3 \cdot \frac{sin ( x )}{cos ( x )} + \frac{cos ( x )}{sin ( x )} - 5 sin (x) : \frac{3 sin ^{2} ( x ) + cos ^{2} ( x ) - 5 sin ^{2} ( x ) cos ( x )}{cos ( x ) sin ( x )}

3 \cdot \frac{sin ( x )}{cos ( x )} + \frac{cos ( x )}{sin ( x )} - 5 sin (x)

乘

3 \cdot \frac{sin ( x )}{cos ( x )} : \frac{3 sin ( x )}{cos ( x )}

3 \cdot \frac{sin ( x )}{cos ( x )}

分式相乘:

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

= \frac{sin ( x ) \cdot 3}{cos ( x )}

= \frac{3 sin ( x )}{cos ( x )} + \frac{cos ( x )}{sin ( x )} - 5 sin (x)

将项转换为分式:

5 sin (x) = \frac{5 sin ( x )}{1}

= \frac{sin ( x ) \cdot 3}{cos ( x )} + \frac{cos ( x )}{sin ( x )} - \frac{5 sin ( x )}{1}

cos (x), sin (x), 1

的最小公倍数:

cos (x) sin (x)

cos (x), sin (x), 1

最小公倍数 (LCM)

计算出由至少在以下一个因式表达式中出现的因子组成的表达式

= cos (x) sin (x)

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) sin (x)

对于

\frac{sin ( x ) \cdot 3}{cos ( x )} :

将分母和分子乘以

sin (x)

\frac{sin ( x ) \cdot 3}{cos ( x )} = \frac{sin ( x ) \cdot 3 sin ( x )}{cos ( x ) sin ( x )} = \frac{3 sin ^{2} ( x )}{cos ( x ) sin ( x )}

对于

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

将分母和分子乘以

cos (x)

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

对于

\frac{5 sin ( x )}{1} :

将分母和分子乘以

cos (x) sin (x)

\frac{5 sin ( x )}{1} = \frac{5 sin ( x ) cos ( x ) sin ( x )}{1 \cdot cos ( x ) sin ( x )} = \frac{5 sin ^{2} ( x ) cos ( x )}{cos ( x ) sin ( x )}

= \frac{3 sin ^{2} ( x )}{cos ( x ) sin ( x )} + \frac{cos ^{2} ( x )}{cos ( x ) sin ( x )} - \frac{5 sin ^{2} ( x ) cos ( x )}{cos ( x ) sin ( 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{3 sin ^{2} ( x ) + cos ^{2} ( x ) - 5 sin ^{2} ( x ) cos ( x )}{cos ( x ) sin ( x )}

\frac{3 sin ^{2} ( x ) + cos ^{2} ( x ) - 5 sin ^{2} ( x ) cos ( x )}{cos ( x ) sin ( x )} < 0

确定

0 \leq x < 2 π

时

\frac{3 sin ^{2} ( x ) + cos ^{2} ( x ) - 5 sin ^{2} ( x ) cos ( x )}{cos ( x ) sin ( x )}

的零点和无定义点

要找到零点，将不等式设置为零

\frac{3 sin ^{2} ( x ) + cos ^{2} ( x ) - 5 sin ^{2} ( x ) cos ( x )}{cos ( x ) sin ( x )} = 0

\frac{3 sin ^{2} ( x ) + cos ^{2} ( x ) - 5 sin ^{2} ( x ) cos ( x )}{cos ( x ) sin ( x )} = 0, 0 \leq x < 2 π : x \in R

无解

\frac{3 sin ^{2} ( x ) + cos ^{2} ( x ) - 5 sin ^{2} ( x ) cos ( x )}{cos ( x ) sin ( x )} = 0, 0 \leq x < 2 π

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

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

使用三角恒等式改写

cos^{2} (x) + 3 sin^{2} (x) - 5 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) + 3 (1 - cos^{2} (x)) - 5 cos (x) (1 - cos^{2} (x))

化简

cos^{2} (x) + 3 (1 - cos^{2} (x)) - 5 cos (x) (1 - cos^{2} (x)) : - 2 cos^{2} (x) - 5 cos (x) + 5 cos^{3} (x) + 3

cos^{2} (x) + 3 (1 - cos^{2} (x)) - 5 cos (x) (1 - cos^{2} (x))

乘开

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

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

使用分配律:

a (b - c) = ab - a c

a = 3, b = 1, c = cos^{2} (x)

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

数字相乘:

3 \cdot 1 = 3

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

= cos^{2} (x) + 3 - 3 cos^{2} (x) - 5 cos (x) (1 - cos^{2} (x))

乘开

- 5 cos (x) (1 - cos^{2} (x)) : - 5 cos (x) + 5 cos^{3} (x)

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

使用分配律:

a (b - c) = ab - a c

a = - 5 cos (x), b = 1, c = cos^{2} (x)

= - 5 cos (x) \cdot 1 - (- 5 cos (x)) cos^{2} (x)

使用加减运算法则

- (- a) = a

= - 5 \cdot 1 \cdot cos (x) + 5 cos^{2} (x) cos (x)

化简

- 5 \cdot 1 \cdot cos (x) + 5 cos^{2} (x) cos (x) : - 5 cos (x) + 5 cos^{3} (x)

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

5 \cdot 1 \cdot cos (x) = 5 cos (x)

5 \cdot 1 \cdot cos (x)

数字相乘:

5 \cdot 1 = 5

= 5 cos (x)

5 cos^{2} (x) cos (x) = 5 cos^{3} (x)

5 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)

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

数字相加:

2 + 1 = 3

= 5 cos^{3} (x)

= - 5 cos (x) + 5 cos^{3} (x)

= - 5 cos (x) + 5 cos^{3} (x)

= cos^{2} (x) + 3 - 3 cos^{2} (x) - 5 cos (x) + 5 cos^{3} (x)

化简

cos^{2} (x) + 3 - 3 cos^{2} (x) - 5 cos (x) + 5 cos^{3} (x) : - 2 cos^{2} (x) - 5 cos (x) + 5 cos^{3} (x) + 3

cos^{2} (x) + 3 - 3 cos^{2} (x) - 5 cos (x) + 5 cos^{3} (x)

对同类项分组

= cos^{2} (x) - 3 cos^{2} (x) - 5 cos (x) + 5 cos^{3} (x) + 3

同类项相加:

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

= - 2 cos^{2} (x) - 5 cos (x) + 5 cos^{3} (x) + 3

= - 2 cos^{2} (x) - 5 cos (x) + 5 cos^{3} (x) + 3

= - 2 cos^{2} (x) - 5 cos (x) + 5 cos^{3} (x) + 3

3 - 2 cos^{2} (x) - 5 cos (x) + 5 cos^{3} (x) = 0

用替代法求解

3 - 2 cos^{2} (x) - 5 cos (x) + 5 cos^{3} (x) = 0

令

: cos (x) = u

3 - 2 u^{2} - 5 u + 5 u^{3} = 0

3 - 2 u^{2} - 5 u + 5 u^{3} = 0 : u \approx - 1.06603 \dots

3 - 2 u^{2} - 5 u + 5 u^{3} = 0

改写成标准形式

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

5 u^{3} - 2 u^{2} - 5 u + 3 = 0

使用牛顿-拉弗森方法找到

5 u^{3} - 2 u^{2} - 5 u + 3 = 0

的一个解:

u \approx - 1.06603 \dots

5 u^{3} - 2 u^{2} - 5 u + 3 = 0

牛顿-拉弗森近似法定义

f (u) = 5 u^{3} - 2 u^{2} - 5 u + 3

找到

f^{^{'}} (u) : 15 u^{2} - 4 u - 5

\frac{d}{d u} (5 u^{3} - 2 u^{2} - 5 u + 3)

使用微分加减法定则:

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

= \frac{d}{d u} (5 u^{3}) - \frac{d}{d u} (2 u^{2}) - \frac{d}{d u} (5 u) + \frac{d}{d u} (3)

\frac{d}{d u} (5 u^{3}) = 15 u^{2}

\frac{d}{d u} (5 u^{3})

将常数提出:

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

= 5 \frac{d}{d u} (u^{3})

使用幂法则:

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

= 5 \cdot 3 u^{3 - 1}

化简

= 15 u^{2}

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

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

将常数提出:

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

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

使用幂法则:

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

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

化简

= 4 u

\frac{d}{d u} (5 u) = 5

\frac{d}{d u} (5 u)

将常数提出:

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

= 5 \frac{d u}{d u}

使用常见微分定则:

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

= 5 \cdot 1

化简

= 5

\frac{d}{d u} (3) = 0

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

常数微分:

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

= 0

= 15 u^{2} - 4 u - 5 + 0

化简

= 15 u^{2} - 4 u - 5

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}) = 5 (- 1)^{3} - 2 (- 1)^{2} - 5 (- 1) + 3 = 1

f^{^{'}} (u_{0}) = 15 (- 1)^{2} - 4 (- 1) - 5 = 14

u_{1} = - 1.07142 \dots

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

Δ u_{1} = 0.07142 \dots

u_{2} = - 1.06606 \dots : Δ u_{2} = 0.00536 \dots

f (u_{1}) = 5 (- 1.07142 \dots)^{3} - 2 (- 1.07142 \dots)^{2} - 5 (- 1.07142 \dots) + 3 = - 0.08855 \dots

f^{^{'}} (u_{1}) = 15 (- 1.07142 \dots)^{2} - 4 (- 1.07142 \dots) - 5 = 16.50510 \dots

u_{2} = - 1.06606 \dots

Δ u_{2} = ∣ - 1.06606 \dots - (- 1.07142 \dots) ∣ = 0.00536 \dots

Δ u_{2} = 0.00536 \dots

u_{3} = - 1.06603 \dots : Δ u_{3} = 0.00003 \dots

f (u_{2}) = 5 (- 1.06606 \dots)^{3} - 2 (- 1.06606 \dots)^{2} - 5 (- 1.06606 \dots) + 3 = - 0.00051 \dots

f^{^{'}} (u_{2}) = 15 (- 1.06606 \dots)^{2} - 4 (- 1.06606 \dots) - 5 = 16.31161 \dots

u_{3} = - 1.06603 \dots

Δ u_{3} = ∣ - 1.06603 \dots - (- 1.06606 \dots) ∣ = 0.00003 \dots

Δ u_{3} = 0.00003 \dots

u_{4} = - 1.06603 \dots : Δ u_{4} = 1.11867 E - 9

f (u_{3}) = 5 (- 1.06603 \dots)^{3} - 2 (- 1.06603 \dots)^{2} - 5 (- 1.06603 \dots) + 3 = - 1.8246 E - 8

f^{^{'}} (u_{3}) = 15 (- 1.06603 \dots)^{2} - 4 (- 1.06603 \dots) - 5 = 16.31046 \dots

u_{4} = - 1.06603 \dots

Δ u_{4} = ∣ - 1.06603 \dots - (- 1.06603 \dots) ∣ = 1.11867 E - 9

Δ u_{4} = 1.11867 E - 9

u \approx - 1.06603 \dots

使用长除法

Eq u a t i o n 0 : \frac{5 u ^{3} - 2 u ^{2} - 5 u + 3}{u + 1.06603 \dots} = 5 u^{2} - 7.33015 \dots u + 2.81417 \dots

5 u^{2} - 7.33015 \dots u + 2.81417 \dots \approx 0

使用牛顿-拉弗森方法找到

5 u^{2} - 7.33015 \dots u + 2.81417 \dots = 0

的一个解:

u \in R

无解

5 u^{2} - 7.33015 \dots u + 2.81417 \dots = 0

牛顿-拉弗森近似法定义

f (u) = 5 u^{2} - 7.33015 \dots u + 2.81417 \dots

找到

f^{^{'}} (u) : 10 u - 7.33015 \dots

\frac{d}{d u} (5 u^{2} - 7.33015 \dots u + 2.81417 \dots)

使用微分加减法定则:

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

= \frac{d}{d u} (5 u^{2}) - \frac{d}{d u} (7.33015 \dots u) + \frac{d}{d u} (2.81417 \dots)

\frac{d}{d u} (5 u^{2}) = 10 u

\frac{d}{d u} (5 u^{2})

将常数提出:

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

= 5 \frac{d}{d u} (u^{2})

使用幂法则:

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

= 5 \cdot 2 u^{2 - 1}

化简

= 10 u

\frac{d}{d u} (7.33015 \dots u) = 7.33015 \dots

\frac{d}{d u} (7.33015 \dots u)

将常数提出:

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

= 7.33015 \dots \frac{d u}{d u}

使用常见微分定则:

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

= 7.33015 \dots \cdot 1

化简

= 7.33015 \dots

\frac{d}{d u} (2.81417 \dots) = 0

\frac{d}{d u} (2.81417 \dots)

常数微分:

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

= 0

= 10 u - 7.33015 \dots + 0

化简

= 10 u - 7.33015 \dots

令

u_{0} = 0

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 0.38391 \dots : Δ u_{1} = 0.38391 \dots

f (u_{0}) = 5 \cdot 0^{2} - 7.33015 \dots \cdot 0 + 2.81417 \dots = 2.81417 \dots

f^{^{'}} (u_{0}) = 10 \cdot 0 - 7.33015 \dots = - 7.33015 \dots

u_{1} = 0.38391 \dots

Δ u_{1} = ∣ 0.38391 \dots - 0 ∣ = 0.38391 \dots

Δ u_{1} = 0.38391 \dots

u_{2} = 0.59502 \dots : Δ u_{2} = 0.21110 \dots

f (u_{1}) = 5 \cdot 0.38391 \dots^{2} - 7.33015 \dots \cdot 0.38391 \dots + 2.81417 \dots = 0.73696 \dots

f^{^{'}} (u_{1}) = 10 \cdot 0.38391 \dots - 7.33015 \dots = - 3.49098 \dots

u_{2} = 0.59502 \dots

Δ u_{2} = ∣ 0.59502 \dots - 0.38391 \dots ∣ = 0.21110 \dots

Δ u_{2} = 0.21110 \dots

u_{3} = 0.75649 \dots : Δ u_{3} = 0.16147 \dots

f (u_{2}) = 5 \cdot 0.59502 \dots^{2} - 7.33015 \dots \cdot 0.59502 \dots + 2.81417 \dots = 0.22282 \dots

f^{^{'}} (u_{2}) = 10 \cdot 0.59502 \dots - 7.33015 \dots = - 1.37992 \dots

u_{3} = 0.75649 \dots

Δ u_{3} = ∣ 0.75649 \dots - 0.59502 \dots ∣ = 0.16147 \dots

Δ u_{3} = 0.16147 \dots

u_{4} = 0.20133 \dots : Δ u_{4} = 0.55516 \dots

f (u_{3}) = 5 \cdot 0.75649 \dots^{2} - 7.33015 \dots \cdot 0.75649 \dots + 2.81417 \dots = 0.13037 \dots

f^{^{'}} (u_{3}) = 10 \cdot 0.75649 \dots - 7.33015 \dots = 0.23484 \dots

u_{4} = 0.20133 \dots

Δ u_{4} = ∣ 0.20133 \dots - 0.75649 \dots ∣ = 0.55516 \dots

Δ u_{4} = 0.55516 \dots

u_{5} = 0.49118 \dots : Δ u_{5} = 0.28984 \dots

f (u_{4}) = 5 \cdot 0.20133 \dots^{2} - 7.33015 \dots \cdot 0.20133 \dots + 2.81417 \dots = 1.54101 \dots

f^{^{'}} (u_{4}) = 10 \cdot 0.20133 \dots - 7.33015 \dots = - 5.31676 \dots

u_{5} = 0.49118 \dots

Δ u_{5} = ∣ 0.49118 \dots - 0.20133 \dots ∣ = 0.28984 \dots

Δ u_{5} = 0.28984 \dots

u_{6} = 0.66486 \dots : Δ u_{6} = 0.17368 \dots

f (u_{5}) = 5 \cdot 0.49118 \dots^{2} - 7.33015 \dots \cdot 0.49118 \dots + 2.81417 \dots = 0.42003 \dots

f^{^{'}} (u_{5}) = 10 \cdot 0.49118 \dots - 7.33015 \dots = - 2.41835 \dots

u_{6} = 0.66486 \dots

Δ u_{6} = ∣ 0.66486 \dots - 0.49118 \dots ∣ = 0.17368 \dots

Δ u_{6} = 0.17368 \dots

u_{7} = 0.88620 \dots : Δ u_{7} = 0.22133 \dots

f (u_{6}) = 5 \cdot 0.66486 \dots^{2} - 7.33015 \dots \cdot 0.66486 \dots + 2.81417 \dots = 0.15083 \dots

f^{^{'}} (u_{6}) = 10 \cdot 0.66486 \dots - 7.33015 \dots = - 0.68147 \dots

u_{7} = 0.88620 \dots

Δ u_{7} = ∣ 0.88620 \dots - 0.66486 \dots ∣ = 0.22133 \dots

Δ u_{7} = 0.22133 \dots

u_{8} = 0.72630 \dots : Δ u_{8} = 0.15990 \dots

f (u_{7}) = 5 \cdot 0.88620 \dots^{2} - 7.33015 \dots \cdot 0.88620 \dots + 2.81417 \dots = 0.24495 \dots

f^{^{'}} (u_{7}) = 10 \cdot 0.88620 \dots - 7.33015 \dots = 1.53190 \dots

u_{8} = 0.72630 \dots

Δ u_{8} = ∣ 0.72630 \dots - 0.88620 \dots ∣ = 0.15990 \dots

Δ u_{8} = 0.15990 \dots

u_{9} = 2.63145 \dots : Δ u_{9} = 1.90514 \dots

f (u_{8}) = 5 \cdot 0.72630 \dots^{2} - 7.33015 \dots \cdot 0.72630 \dots + 2.81417 \dots = 0.12784 \dots

f^{^{'}} (u_{8}) = 10 \cdot 0.72630 \dots - 7.33015 \dots = - 0.06710 \dots

u_{9} = 2.63145 \dots

Δ u_{9} = ∣ 2.63145 \dots - 0.72630 \dots ∣ = 1.90514 \dots

Δ u_{9} = 1.90514 \dots

u_{10} = 1.67551 \dots : Δ u_{10} = 0.95594 \dots

f (u_{9}) = 5 \cdot 2.63145 \dots^{2} - 7.33015 \dots \cdot 2.63145 \dots + 2.81417 \dots = 18.14798 \dots

f^{^{'}} (u_{9}) = 10 \cdot 2.63145 \dots - 7.33015 \dots = 18.98439 \dots

u_{10} = 1.67551 \dots

Δ u_{10} = ∣ 1.67551 \dots - 2.63145 \dots ∣ = 0.95594 \dots

Δ u_{10} = 0.95594 \dots

u_{11} = 1.19072 \dots : Δ u_{11} = 0.48478 \dots

f (u_{10}) = 5 \cdot 1.67551 \dots^{2} - 7.33015 \dots \cdot 1.67551 \dots + 2.81417 \dots = 4.56912 \dots

f^{^{'}} (u_{10}) = 10 \cdot 1.67551 \dots - 7.33015 \dots = 9.42497 \dots

u_{11} = 1.19072 \dots

Δ u_{11} = ∣ 1.19072 \dots - 1.67551 \dots ∣ = 0.48478 \dots

Δ u_{11} = 0.48478 \dots

u_{12} = 0.93398 \dots : Δ u_{12} = 0.25673 \dots

f (u_{11}) = 5 \cdot 1.19072 \dots^{2} - 7.33015 \dots \cdot 1.19072 \dots + 2.81417 \dots = 1.17510 \dots

f^{^{'}} (u_{11}) = 10 \cdot 1.19072 \dots - 7.33015 \dots = 4.57708 \dots

u_{12} = 0.93398 \dots

Δ u_{12} = ∣ 0.93398 \dots - 1.19072 \dots ∣ = 0.25673 \dots

Δ u_{12} = 0.25673 \dots

u_{13} = 0.77000 \dots : Δ u_{13} = 0.16398 \dots

f (u_{12}) = 5 \cdot 0.93398 \dots^{2} - 7.33015 \dots \cdot 0.93398 \dots + 2.81417 \dots = 0.32956 \dots

f^{^{'}} (u_{12}) = 10 \cdot 0.93398 \dots - 7.33015 \dots = 2.00972 \dots

u_{13} = 0.77000 \dots

Δ u_{13} = ∣ 0.77000 \dots - 0.93398 \dots ∣ = 0.16398 \dots

Δ u_{13} = 0.16398 \dots

u_{14} = 0.40647 \dots : Δ u_{14} = 0.36352 \dots

f (u_{13}) = 5 \cdot 0.77000 \dots^{2} - 7.33015 \dots \cdot 0.77000 \dots + 2.81417 \dots = 0.13445 \dots

f^{^{'}} (u_{13}) = 10 \cdot 0.77000 \dots - 7.33015 \dots = 0.36986 \dots

u_{14} = 0.40647 \dots

Δ u_{14} = ∣ 0.40647 \dots - 0.77000 \dots ∣ = 0.36352 \dots

Δ u_{14} = 0.36352 \dots

无法得出解

解是

u \approx - 1.06603 \dots

u = cos (x)

代回

cos (x) \approx - 1.06603 \dots

cos (x) \approx - 1.06603 \dots

cos (x) = - 1.06603 \dots, 0 \leq x < 2 π :

无解

cos (x) = - 1.06603 \dots, 0 \leq x < 2 π

- 1 \leq cos (x) \leq 1

无解

合并所有解

x \in R 无解

确定无定义点:

x = \frac{π}{2}, x = \frac{3 π}{2}, x = 0, x = π

找到分母的零解

cos (x) sin (x) = 0

分别求解每个部分

cos (x) = 0 or sin (x) = 0

cos (x) = 0, 0 \leq x < 2 π : x = \frac{π}{2}, x = \frac{3 π}{2}

cos (x) = 0, 0 \leq x < 2 π

cos (x) = 0

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 = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

在

0 \leq x < 2 π

范围内的解

x = \frac{π}{2}, x = \frac{3 π}{2}

sin (x) = 0, 0 \leq x < 2 π : x = 0, x = π

sin (x) = 0, 0 \leq x < 2 π

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

在

0 \leq x < 2 π

范围内的解

x = 0, x = π

合并所有解

x = \frac{π}{2}, x = \frac{3 π}{2}, x = 0, x = π

0, \frac{π}{2}, π, \frac{3 π}{2}

确定区间

0 < x < \frac{π}{2}, \frac{π}{2} < x < π, π < x < \frac{3 π}{2}, \frac{3 π}{2} < x < 2 π

总结如下表：

3 sin^{2} (x) + cos^{2} (x) - 5 sin^{2} (x) cos (x) cos (x) sin (x) \frac{3 s i n ^{2} ( x ) + c o s ^{2} ( x ) - 5 s i n ^{2} ( x ) c o s ( x )}{c o s ( x ) s i n ( x )} x = 0 + + 0 未定义 0 < x < \frac{π}{2} + + + + x = \frac{π}{2} + 0 + 未定义 \frac{π}{2} < x < π + - + - x = π + - 0 未定义 π < x < \frac{3 π}{2} + - - + x = \frac{3 π}{2} + 0 - 未定义 \frac{3 π}{2} < x < 2 π + + - - x = 2 π + + 0 未定义

确定满足所需条件的区间：

< 0

\frac{π}{2} < x < π or \frac{3 π}{2} < x < 2 π

使用周期

3 tan (x) + cot (x) - 5 sin (x)

\frac{π}{2} + 2 πn < x < π + 2 πn or \frac{3 π}{2} + 2 πn < x < 2 π + 2 πn

流行的例子

sin(x)-1/2 sqrt(3)<0

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

2cos(x)>cos(2x)

2 cos (x) > cos (2 x)

tan^{2} (x) < 1

sin(2t)<1,(0,2pi)

sin (2 t) < 1, (0, 2 π)

2sin^2(x)-3sin(x)+1<= 0

2 sin^{2} (x) - 3 sin (x) + 1 \leq 0