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受欢迎的三角函数 >

2csc^2(θ)-3cot(θ)+3=0

解答

2 csc^{2} (θ) - 3 cot (θ) + 3 = 0

解答

θ \in R 无解

求解步骤

2 csc^{2} (θ) - 3 cot (θ) + 3 = 0

使用三角恒等式改写

3 + 2 csc^{2} (θ) - 3 cot (θ)

使用毕达哥拉斯恒等式:

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

= 3 + 2 (1 + cot^{2} (θ)) - 3 cot (θ)

化简

3 + 2 (1 + cot^{2} (θ)) - 3 cot (θ) : 2 cot^{2} (θ) - 3 cot (θ) + 5

3 + 2 (1 + cot^{2} (θ)) - 3 cot (θ)

乘开

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

2 (1 + cot^{2} (θ))

使用分配律:

a (b + c) = ab + a c

a = 2, b = 1, c = cot^{2} (θ)

= 2 \cdot 1 + 2 cot^{2} (θ)

数字相乘:

2 \cdot 1 = 2

= 2 + 2 cot^{2} (θ)

= 3 + 2 + 2 cot^{2} (θ) - 3 cot (θ)

数字相加:

3 + 2 = 5

= 2 cot^{2} (θ) - 3 cot (θ) + 5

= 2 cot^{2} (θ) - 3 cot (θ) + 5

5 + 2 cot^{2} (θ) - 3 cot (θ) = 0

用替代法求解

5 + 2 cot^{2} (θ) - 3 cot (θ) = 0

令

: cot (θ) = u

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

5 + 2 u^{2} - 3 u = 0 : u = \frac{3}{4} + i \frac{31}{4}, u = \frac{3}{4} - i \frac{31}{4}

5 + 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 + 5 = 0

使用求根公式求解

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

二次方程求根公式:

若

a = 2, b = - 3, c = 5

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 \cdot 5}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 \cdot 5}{2 \cdot 2}

化简

(- 3)^{2} - 4 \cdot 2 \cdot 5 : 31 i

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

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

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

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

数字相乘:

4 \cdot 2 \cdot 5 = 40

= 3^{2} - 40

使用虚数运算法则:

- a = i a

= i 40 - 3^{2}

- 3^{2} + 40 = 31

- 3^{2} + 40

3^{2} = 9

= - 9 + 40

数字相加/相减:

- 9 + 40 = 31

= 31

= 31 i

u_{1, 2} = \frac{- ( - 3 ) \pm 31 i}{2 \cdot 2}

将解分隔开

u_{1} = \frac{- ( - 3 ) + 31 i}{2 \cdot 2}, u_{2} = \frac{- ( - 3 ) - 31 i}{2 \cdot 2}

u = \frac{- ( - 3 ) + 31 i}{2 \cdot 2} : \frac{3}{4} + i \frac{31}{4}

\frac{- ( - 3 ) + 31 i}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{3 + 31 i}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{3 + 31 i}{4}

将

\frac{3 + 31 i}{4}

改写成标准复数形式:

\frac{3}{4} + \frac{31}{4} i

\frac{3 + 31 i}{4}

使用分式法则:

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

\frac{3 + 31 i}{4} = \frac{3}{4} + \frac{31 i}{4}

= \frac{3}{4} + \frac{31 i}{4}

= \frac{3}{4} + \frac{31}{4} i

u = \frac{- ( - 3 ) - 31 i}{2 \cdot 2} : \frac{3}{4} - i \frac{31}{4}

\frac{- ( - 3 ) - 31 i}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{3 - 31 i}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{3 - 31 i}{4}

将

\frac{3 - 31 i}{4}

改写成标准复数形式:

\frac{3}{4} - \frac{31}{4} i

\frac{3 - 31 i}{4}

使用分式法则:

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

\frac{3 - 31 i}{4} = \frac{3}{4} - \frac{31 i}{4}

= \frac{3}{4} - \frac{31 i}{4}

= \frac{3}{4} - \frac{31}{4} i

二次方程组的解是:

u = \frac{3}{4} + i \frac{31}{4}, u = \frac{3}{4} - i \frac{31}{4}

u = cot (θ)

代回

cot (θ) = \frac{3}{4} + i \frac{31}{4}, cot (θ) = \frac{3}{4} - i \frac{31}{4}

cot (θ) = \frac{3}{4} + i \frac{31}{4}, cot (θ) = \frac{3}{4} - i \frac{31}{4}

cot (θ) = \frac{3}{4} + i \frac{31}{4} :

无解

cot (θ) = \frac{3}{4} + i \frac{31}{4}

无解

cot (θ) = \frac{3}{4} - i \frac{31}{4} :

无解

cot (θ) = \frac{3}{4} - i \frac{31}{4}

无解

合并所有解

θ \in R 无解

作图

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