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 >

cot^2(x)+csc(x)=1,0<= x<2pi

解答

cot^{2} (x) + csc (x) = 1, 0 \leq x < 2 π

解答

x = \frac{π}{2}, x = \frac{7 π}{6}, x = \frac{11 π}{6}

+1

度数

x = 9 0^{\circ}, x = 21 0^{\circ}, x = 33 0^{\circ}

求解步骤

cot^{2} (x) + csc (x) = 1, 0 \leq x < 2 π

两边减去

1

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

使用三角恒等式改写

- 1 + cot^{2} (x) + csc (x)

使用毕达哥拉斯恒等式:

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

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

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

化简

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

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

对同类项分组

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

数字相减:

- 1 - 1 = - 2

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

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

- 2 + csc (x) + csc^{2} (x) = 0

用替代法求解

- 2 + csc (x) + csc^{2} (x) = 0

令

: csc (x) = u

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

- 2 + u + u^{2} = 0 : u = 1, u = - 2

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

改写成标准形式

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

u^{2} + u - 2 = 0

使用求根公式求解

u^{2} + u - 2 = 0

二次方程求根公式:

若

a = 1, b = 1, c = - 2

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

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

1^{2} - 4 \cdot 1 \cdot (- 2) = 3

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 2)

使用法则

- (- a) = a

= 1 + 4 \cdot 1 \cdot 2

数字相乘:

4 \cdot 1 \cdot 2 = 8

= 1 + 8

数字相加:

1 + 8 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

3^{2} = 3

= 3

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

将解分隔开

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

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

\frac{- 1 + 3}{2 \cdot 1}

数字相加/相减:

- 1 + 3 = 2

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

数字相乘:

2 \cdot 1 = 2

= \frac{2}{2}

使用法则

\frac{a}{a} = 1

= 1

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

\frac{- 1 - 3}{2 \cdot 1}

数字相减:

- 1 - 3 = - 4

= \frac{- 4}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 4}{2}

使用分式法则:

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

= - \frac{4}{2}

数字相除:

\frac{4}{2} = 2

= - 2

二次方程组的解是:

u = 1, u = - 2

u = csc (x)

代回

csc (x) = 1, csc (x) = - 2

csc (x) = 1, csc (x) = - 2

csc (x) = 1, 0 \leq x < 2 π : x = \frac{π}{2}

csc (x) = 1, 0 \leq x < 2 π

csc (x) = 1

的通解

csc (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} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 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} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

在

0 \leq x < 2 π

范围内的解

x = \frac{π}{2}

csc (x) = - 2, 0 \leq x < 2 π : x = \frac{7 π}{6}, x = \frac{11 π}{6}

csc (x) = - 2, 0 \leq x < 2 π

csc (x) = - 2

的通解

csc (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} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 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} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

在

0 \leq x < 2 π

范围内的解

x = \frac{7 π}{6}, x = \frac{11 π}{6}

合并所有解

x = \frac{π}{2}, x = \frac{7 π}{6}, x = \frac{11 π}{6}

作图

流行的例子

81pi=10.5-9cos(pi/(30)t)

cot(x)+tan(x)=1

2cos(x)-3sin(x)=-2