zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

csc(x)cot(x)= 2/3

解答

csc (x) cot (x) = \frac{2}{3}

解答

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

+1

度数

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

求解步骤

csc (x) cot (x) = \frac{2}{3}

两边减去

\frac{2}{3}

csc (x) cot (x) - \frac{2}{3} = 0

化简

csc (x) cot (x) - \frac{2}{3} : \frac{3 csc ( x ) cot ( x ) - 2}{3}

csc (x) cot (x) - \frac{2}{3}

将项转换为分式:

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

= \frac{csc ( x ) cot ( x ) \cdot 3}{3} - \frac{2}{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 ) cot ( x ) \cdot 3 - 2}{3}

\frac{3 csc ( x ) cot ( x ) - 2}{3} = 0

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

3 csc (x) cot (x) - 2 = 0

用 sin, cos 表示

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

化简

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

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

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

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

分式相乘:

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

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

数字相乘:

1 \cdot 3 = 3

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

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

sin (x) sin (x)

使用指数法则:

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

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

= sin^{1 + 1} (x)

数字相加:

1 + 1 = 2

= sin^{2} (x)

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

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

将项转换为分式:

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

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

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

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

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

两边加上

2 sin^{2} (x)

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

两边进行平方

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

两边减去

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

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

分解

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

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

将

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

改写为

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

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

将

9

改写为

3^{2}

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

将

4

改写为

2^{2}

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

使用指数法则:

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

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

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

使用指数法则:

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

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

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

使用指数法则:

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

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

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

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

使用平方差公式:

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

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

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

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

分别求解每个部分

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

3 cos (x) + 2 sin^{2} (x) = 0 : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

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

使用三角恒等式改写

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

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

(1 - cos^{2} (x)) \cdot 2 + 3 cos (x) = 0

用替代法求解

(1 - cos^{2} (x)) \cdot 2 + 3 cos (x) = 0

令

: cos (x) = u

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

(1 - u^{2}) \cdot 2 + 3 u = 0 : u = - \frac{1}{2}, u = 2

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

展开

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

(1 - u^{2}) \cdot 2 + 3 u

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

乘开

2 (1 - u^{2}) : 2 - 2 u^{2}

2 (1 - u^{2})

使用分配律:

a (b - c) = ab - a c

a = 2, b = 1, c = u^{2}

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

数字相乘:

2 \cdot 1 = 2

= 2 - 2 u^{2}

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = - 2, b = 3, c = 2

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 2 ) \cdot 2}{2 ( - 2 )}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 2 ) \cdot 2}{2 ( - 2 )}

3^{2} - 4 (- 2) \cdot 2 = 5

3^{2} - 4 (- 2) \cdot 2

使用法则

- (- a) = a

= 3^{2} + 4 \cdot 2 \cdot 2

数字相乘:

4 \cdot 2 \cdot 2 = 16

= 3^{2} + 16

3^{2} = 9

= 9 + 16

数字相加:

9 + 16 = 25

= 25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 5

u_{1, 2} = \frac{- 3 \pm 5}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- 3 + 5}{2 ( - 2 )}, u_{2} = \frac{- 3 - 5}{2 ( - 2 )}

u = \frac{- 3 + 5}{2 ( - 2 )} : - \frac{1}{2}

\frac{- 3 + 5}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 3 + 5}{- 2 \cdot 2}

数字相加/相减:

- 3 + 5 = 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{2}{- 4}

使用分式法则:

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

= - \frac{2}{4}

约分:

2

= - \frac{1}{2}

u = \frac{- 3 - 5}{2 ( - 2 )} : 2

\frac{- 3 - 5}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 3 - 5}{- 2 \cdot 2}

数字相减:

- 3 - 5 = - 8

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 8}{- 4}

使用分式法则:

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

= \frac{8}{4}

数字相除:

\frac{8}{4} = 2

= 2

二次方程组的解是:

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

u = cos (x)

代回

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

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

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) = 2 :

无解

cos (x) = 2

- 1 \leq cos (x) \leq 1

无解

合并所有解

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

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

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

使用三角恒等式改写

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

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

- (1 - cos^{2} (x)) \cdot 2 + 3 cos (x) = 0

用替代法求解

- (1 - cos^{2} (x)) \cdot 2 + 3 cos (x) = 0

令

: cos (x) = u

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

- (1 - u^{2}) \cdot 2 + 3 u = 0 : u = \frac{1}{2}, u = - 2

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

展开

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

- (1 - u^{2}) \cdot 2 + 3 u

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

乘开

- 2 (1 - u^{2}) : - 2 + 2 u^{2}

- 2 (1 - u^{2})

使用分配律:

a (b - c) = ab - a c

a = - 2, b = 1, c = u^{2}

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

使用加减运算法则

- (- a) = a

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

数字相乘:

2 \cdot 1 = 2

= - 2 + 2 u^{2}

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = 2, b = 3, c = - 2

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 ( - 2 )}{2 \cdot 2}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 ( - 2 )}{2 \cdot 2}

3^{2} - 4 \cdot 2 (- 2) = 5

3^{2} - 4 \cdot 2 (- 2)

使用法则

- (- a) = a

= 3^{2} + 4 \cdot 2 \cdot 2

数字相乘:

4 \cdot 2 \cdot 2 = 16

= 3^{2} + 16

3^{2} = 9

= 9 + 16

数字相加:

9 + 16 = 25

= 25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 5

u_{1, 2} = \frac{- 3 \pm 5}{2 \cdot 2}

将解分隔开

u_{1} = \frac{- 3 + 5}{2 \cdot 2}, u_{2} = \frac{- 3 - 5}{2 \cdot 2}

u = \frac{- 3 + 5}{2 \cdot 2} : \frac{1}{2}

\frac{- 3 + 5}{2 \cdot 2}

数字相加/相减:

- 3 + 5 = 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{2}{4}

约分:

2

= \frac{1}{2}

u = \frac{- 3 - 5}{2 \cdot 2} : - 2

\frac{- 3 - 5}{2 \cdot 2}

数字相减:

- 3 - 5 = - 8

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 8}{4}

使用分式法则:

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

= - \frac{8}{4}

数字相除:

\frac{8}{4} = 2

= - 2

二次方程组的解是:

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

u = cos (x)

代回

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

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

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 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{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = - 2 :

无解

cos (x) = - 2

- 1 \leq cos (x) \leq 1

无解

合并所有解

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

合并所有解

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

将解代入原方程进行验证

将它们代入

csc (x) cot (x) = \frac{2}{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:

假

\frac{2 π}{3} + 2 πn

代入

n = 1

\frac{2 π}{3} + 2 π 1

对于

csc (x) cot (x) = \frac{2}{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{2}{3}

整理后得

- 0.66666 \dots = 0.66666 \dots

\Rightarrow 假

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{2}{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{2}{3}

整理后得

- 0.66666 \dots = 0.66666 \dots

\Rightarrow 假

检验

\frac{π}{3} + 2 πn

的解:

真

\frac{π}{3} + 2 πn

代入

n = 1

\frac{π}{3} + 2 π 1

对于

csc (x) cot (x) = \frac{2}{3}

代入

x = \frac{π}{3} + 2 π 1

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

整理后得

0.66666 \dots = 0.66666 \dots

\Rightarrow 真

检验

\frac{5 π}{3} + 2 πn

的解:

真

\frac{5 π}{3} + 2 πn

代入

n = 1

\frac{5 π}{3} + 2 π 1

对于

csc (x) cot (x) = \frac{2}{3}

代入

x = \frac{5 π}{3} + 2 π 1

csc (\frac{5 π}{3} + 2 π 1) cot (\frac{5 π}{3} + 2 π 1) = \frac{2}{3}

整理后得

0.66666 \dots = 0.66666 \dots

\Rightarrow 真

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

作图

流行的例子

4tan^2(x)-8tan(x)+3=0

4 tan^{2} (x) - 8 tan (x) + 3 = 0

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

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

sin (\frac{3 x}{2}) = 0

tan (x) = 5 sin (x)

cot^2(x)=-2cot(x)-1

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