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受欢迎的三角函数 >

csc^2(θ)+cot(θ)=-2

解答

csc^{2} (θ) + cot (θ) = - 2

解答

θ \in R 无解

求解步骤

csc^{2} (θ) + cot (θ) = - 2

两边减去

- 2

csc^{2} (θ) + cot (θ) + 2 = 0

使用三角恒等式改写

2 + cot (θ) + csc^{2} (θ)

使用毕达哥拉斯恒等式:

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

= 2 + cot (θ) + 1 + cot^{2} (θ)

化简

2 + cot (θ) + 1 + cot^{2} (θ) : cot^{2} (θ) + cot (θ) + 3

2 + cot (θ) + 1 + cot^{2} (θ)

对同类项分组

= cot (θ) + cot^{2} (θ) + 2 + 1

数字相加:

2 + 1 = 3

= cot^{2} (θ) + cot (θ) + 3

= cot^{2} (θ) + cot (θ) + 3

3 + cot (θ) + cot^{2} (θ) = 0

用替代法求解

3 + cot (θ) + cot^{2} (θ) = 0

令

: cot (θ) = u

3 + u + u^{2} = 0

3 + u + u^{2} = 0 : u = - \frac{1}{2} + i \frac{11}{2}, u = - \frac{1}{2} - i \frac{11}{2}

3 + u + u^{2} = 0

改写成标准形式

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

u^{2} + u + 3 = 0

使用求根公式求解

u^{2} + u + 3 = 0

二次方程求根公式:

若

a = 1, b = 1, c = 3

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

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

化简

1^{2} - 4 \cdot 1 \cdot 3 : 11 i

1^{2} - 4 \cdot 1 \cdot 3

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot 3

数字相乘:

4 \cdot 1 \cdot 3 = 12

= 1 - 12

数字相减:

1 - 12 = - 11

= - 11

使用根式运算法则:

- a = - 1 a

- 11 = - 1 11

= - 1 11

使用虚数运算法则:

- 1 = i

= 11 i

u_{1, 2} = \frac{- 1 \pm 11 i}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- 1 + 11 i}{2 \cdot 1}, u_{2} = \frac{- 1 - 11 i}{2 \cdot 1}

u = \frac{- 1 + 11 i}{2 \cdot 1} : - \frac{1}{2} + i \frac{11}{2}

\frac{- 1 + 11 i}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 1 + 11 i}{2}

将

\frac{- 1 + 11 i}{2}

改写成标准复数形式:

- \frac{1}{2} + \frac{11}{2} i

\frac{- 1 + 11 i}{2}

使用分式法则:

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

\frac{- 1 + 11 i}{2} = - \frac{1}{2} + \frac{11 i}{2}

= - \frac{1}{2} + \frac{11 i}{2}

= - \frac{1}{2} + \frac{11}{2} i

u = \frac{- 1 - 11 i}{2 \cdot 1} : - \frac{1}{2} - i \frac{11}{2}

\frac{- 1 - 11 i}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 1 - 11 i}{2}

将

\frac{- 1 - 11 i}{2}

改写成标准复数形式:

- \frac{1}{2} - \frac{11}{2} i

\frac{- 1 - 11 i}{2}

使用分式法则:

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

\frac{- 1 - 11 i}{2} = - \frac{1}{2} - \frac{11 i}{2}

= - \frac{1}{2} - \frac{11 i}{2}

= - \frac{1}{2} - \frac{11}{2} i

二次方程组的解是:

u = - \frac{1}{2} + i \frac{11}{2}, u = - \frac{1}{2} - i \frac{11}{2}

u = cot (θ)

代回

cot (θ) = - \frac{1}{2} + i \frac{11}{2}, cot (θ) = - \frac{1}{2} - i \frac{11}{2}

cot (θ) = - \frac{1}{2} + i \frac{11}{2}, cot (θ) = - \frac{1}{2} - i \frac{11}{2}

cot (θ) = - \frac{1}{2} + i \frac{11}{2} :

无解

cot (θ) = - \frac{1}{2} + i \frac{11}{2}

无解

cot (θ) = - \frac{1}{2} - i \frac{11}{2} :

无解

cot (θ) = - \frac{1}{2} - i \frac{11}{2}

无解

合并所有解

θ \in R 无解

作图

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