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受欢迎的三角函数 >

solvefor x,csc(x)+cot(x)=(sqrt(3))/3

解答

求解

x, csc (x) + cot (x) = \frac{3}{3}

解答

x = \frac{2 π}{3} + 2 πn

+1

度数

x = 12 0^{\circ} + 36 0^{\circ} n

求解步骤

csc (x) + cot (x) = \frac{3}{3}

两边减去

\frac{3}{3}

csc (x) + cot (x) - \frac{1}{3} = 0

化简

csc (x) + cot (x) - \frac{1}{3} : \frac{3 csc ( x ) + 3 cot ( x ) - 1}{3}

csc (x) + cot (x) - \frac{1}{3}

将项转换为分式:

csc (x) = \frac{csc ( x ) 3}{3}, cot (x) = \frac{cot ( x ) 3}{3}

= \frac{csc ( x ) 3}{3} + \frac{cot ( x ) 3}{3} - \frac{1}{3}

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{csc ( x ) 3 + cot ( x ) 3 - 1}{3}

\frac{3 csc ( x ) + 3 cot ( x ) - 1}{3} = 0

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

3 csc (x) + 3 cot (x) - 1 = 0

用 sin, cos 表示

3 \frac{1}{sin ( x )} + 3 \frac{cos ( x )}{sin ( x )} - 1 = 0

化简

3 \frac{1}{sin ( x )} + 3 \frac{cos ( x )}{sin ( x )} - 1 : \frac{3 + 3 cos ( x ) - sin ( x )}{sin ( x )}

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

3 \frac{1}{sin ( x )} = \frac{3}{sin ( x )}

3 \frac{1}{sin ( x )}

分式相乘:

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

= \frac{1 \cdot 3}{sin ( x )}

乘以:

1 \cdot 3 = 3

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

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

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

分式相乘:

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

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

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

合并分式

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

使用法则

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

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

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

将项转换为分式:

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

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

乘以:

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

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

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

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

3 + 3 cos (x) - sin (x) = 0

两边加上

sin (x)

3 + 3 cos (x) = sin (x)

两边进行平方

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

两边减去

sin^{2} (x)

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

使用三角恒等式改写

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

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

化简

(3 + cos (x) 3)^{2} - (1 - cos^{2} (x)) : 4 cos^{2} (x) + 6 cos (x) + 2

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

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

(3 + cos (x) 3)^{2} : 3 + 6 cos (x) + 3 cos^{2} (x)

使用完全平方公式:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = 3, b = cos (x) 3

= (3)^{2} + 23 cos (x) 3 + (cos (x) 3)^{2}

化简

(3)^{2} + 23 cos (x) 3 + (cos (x) 3)^{2} : 3 + 6 cos (x) + 3 cos^{2} (x)

(3)^{2} + 23 cos (x) 3 + (cos (x) 3)^{2}

(3)^{2} = 3

(3)^{2}

使用根式运算法则:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

使用指数法则:

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

= 3^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

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

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 3

23 cos (x) 3 = 6 cos (x)

23 cos (x) 3

使用根式运算法则:

a a = a

33 = 3

= 2 \cdot 3 cos (x)

数字相乘:

2 \cdot 3 = 6

= 6 cos (x)

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

(cos (x) 3)^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

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

(3)^{2} : 3

使用根式运算法则:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

使用指数法则:

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

= 3^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

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

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 3

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

= 3 + 6 cos (x) + 3 cos^{2} (x)

= 3 + 6 cos (x) + 3 cos^{2} (x)

= 3 + 6 cos (x) + 3 cos^{2} (x) - (1 - cos^{2} (x))

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

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

打开括号

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

使用加减运算法则

- (- a) = a, - (a) = - a

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

= 3 + 6 cos (x) + 3 cos^{2} (x) - 1 + cos^{2} (x)

化简

3 + 6 cos (x) + 3 cos^{2} (x) - 1 + cos^{2} (x) : 4 cos^{2} (x) + 6 cos (x) + 2

3 + 6 cos (x) + 3 cos^{2} (x) - 1 + cos^{2} (x)

对同类项分组

= 6 cos (x) + 3 cos^{2} (x) + cos^{2} (x) + 3 - 1

同类项相加:

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

= 6 cos (x) + 4 cos^{2} (x) + 3 - 1

数字相加/相减:

3 - 1 = 2

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

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

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

2 + 4 cos^{2} (x) + 6 cos (x) = 0

用替代法求解

2 + 4 cos^{2} (x) + 6 cos (x) = 0

令

: cos (x) = u

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

2 + 4 u^{2} + 6 u = 0 : u = - \frac{1}{2}, u = - 1

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

改写成标准形式

a x^{2} + b x + c = 0

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

使用求根公式求解

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

二次方程求根公式:

若

a = 4, b = 6, c = 2

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 4 \cdot 2}{2 \cdot 4}

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 4 \cdot 2}{2 \cdot 4}

6^{2} - 4 \cdot 4 \cdot 2 = 2

6^{2} - 4 \cdot 4 \cdot 2

数字相乘:

4 \cdot 4 \cdot 2 = 32

= 6^{2} - 32

6^{2} = 36

= 36 - 32

数字相减:

36 - 32 = 4

= 4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

u_{1, 2} = \frac{- 6 \pm 2}{2 \cdot 4}

将解分隔开

u_{1} = \frac{- 6 + 2}{2 \cdot 4}, u_{2} = \frac{- 6 - 2}{2 \cdot 4}

u = \frac{- 6 + 2}{2 \cdot 4} : - \frac{1}{2}

\frac{- 6 + 2}{2 \cdot 4}

数字相加/相减:

- 6 + 2 = - 4

= \frac{- 4}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 4}{8}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{4}{8}

约分:

4

= - \frac{1}{2}

u = \frac{- 6 - 2}{2 \cdot 4} : - 1

\frac{- 6 - 2}{2 \cdot 4}

数字相减:

- 6 - 2 = - 8

= \frac{- 8}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 8}{8}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{8}{8}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

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

u = cos (x)

代回

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

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

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

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

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

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{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

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

合并所有解

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = π + 2 πn

将解代入原方程进行验证

将它们代入

csc (x) + cot (x) = \frac{3}{3}

检验解是否符合

去除与方程不符的解。

检验

\frac{2 π}{3} + 2 πn

的解:

真

\frac{2 π}{3} + 2 πn

代入

n = 1

\frac{2 π}{3} + 2 π 1

对于

csc (x) + cot (x) = \frac{3}{3}

代入

x = \frac{2 π}{3} + 2 π 1

csc (\frac{2 π}{3} + 2 π 1) + cot (\frac{2 π}{3} + 2 π 1) = \frac{3}{3}

整理后得

0.57735 \dots = 0.57735 \dots

\Rightarrow 真

检验

\frac{4 π}{3} + 2 πn

的解:

假

\frac{4 π}{3} + 2 πn

代入

n = 1

\frac{4 π}{3} + 2 π 1

对于

csc (x) + cot (x) = \frac{3}{3}

代入

x = \frac{4 π}{3} + 2 π 1

csc (\frac{4 π}{3} + 2 π 1) + cot (\frac{4 π}{3} + 2 π 1) = \frac{3}{3}

整理后得

- 0.57735 \dots = 0.57735 \dots

\Rightarrow 假

检验

π + 2 πn

的解:

假

π + 2 πn

代入

n = 1

π + 2 π 1

对于

csc (x) + cot (x) = \frac{3}{3}

代入

x = π + 2 π 1

csc (π + 2 π 1) + cot (π + 2 π 1) = \frac{3}{3}

未定义

\Rightarrow 假

x = \frac{2 π}{3} + 2 πn

作图

流行的例子

5cos(θ)+1=8cos(θ)

5 cos (θ) + 1 = 8 cos (θ)

tan (x) = - \frac{12}{35}

2cot^2(x)-2csc(x)=2

2 cot^{2} (x) - 2 csc (x) = 2

6cos^3(x)=6cos(x)

6 cos^{3} (x) = 6 cos (x)

cos(x)-cos(2x)=2cos(x)

cos (x) - cos (2 x) = 2 cos (x)