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受欢迎的三角函数 >

cot^2(θ)+cot(θ)-20=0

解答

cot^{2} (θ) + cot (θ) - 20 = 0

解答

θ = 0.24497 \dots + πn, θ = 2.94419 \dots + πn

+1

度数

θ = 14.03624 \dots^{\circ} + 18 0^{\circ} n, θ = 168.69006 \dots^{\circ} + 18 0^{\circ} n

求解步骤

cot^{2} (θ) + cot (θ) - 20 = 0

用替代法求解

cot^{2} (θ) + cot (θ) - 20 = 0

令

: cot (θ) = u

u^{2} + u - 20 = 0

u^{2} + u - 20 = 0 : u = 4, u = - 5

u^{2} + u - 20 = 0

使用求根公式求解

u^{2} + u - 20 = 0

二次方程求根公式:

若

a = 1, b = 1, c = - 20

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 20 )}{2 \cdot 1}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 20 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 20) = 9

1^{2} - 4 \cdot 1 \cdot (- 20)

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 20)

使用法则

- (- a) = a

= 1 + 4 \cdot 1 \cdot 20

数字相乘:

4 \cdot 1 \cdot 20 = 80

= 1 + 80

数字相加:

1 + 80 = 81

= 81

因式分解数字:

81 = 9^{2}

= 9^{2}

使用根式运算法则:

n a^{n} = a

9^{2} = 9

= 9

u_{1, 2} = \frac{- 1 \pm 9}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- 1 + 9}{2 \cdot 1}, u_{2} = \frac{- 1 - 9}{2 \cdot 1}

u = \frac{- 1 + 9}{2 \cdot 1} : 4

\frac{- 1 + 9}{2 \cdot 1}

数字相加/相减:

- 1 + 9 = 8

= \frac{8}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{8}{2}

数字相除:

\frac{8}{2} = 4

= 4

u = \frac{- 1 - 9}{2 \cdot 1} : - 5

\frac{- 1 - 9}{2 \cdot 1}

数字相减:

- 1 - 9 = - 10

= \frac{- 10}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 10}{2}

使用分式法则:

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

= - \frac{10}{2}

数字相除:

\frac{10}{2} = 5

= - 5

二次方程组的解是:

u = 4, u = - 5

u = cot (θ)

代回

cot (θ) = 4, cot (θ) = - 5

cot (θ) = 4, cot (θ) = - 5

cot (θ) = 4 : θ = arccot (4) + πn

cot (θ) = 4

使用反三角函数性质

cot (θ) = 4

cot (θ) = 4

的通解

cot (x) = a \Rightarrow x = arccot (a) + πn

θ = arccot (4) + πn

θ = arccot (4) + πn

cot (θ) = - 5 : θ = arccot (- 5) + πn

cot (θ) = - 5

使用反三角函数性质

cot (θ) = - 5

cot (θ) = - 5

的通解

cot (x) = - a \Rightarrow x = arccot (- a) + πn

θ = arccot (- 5) + πn

θ = arccot (- 5) + πn

合并所有解

θ = arccot (4) + πn, θ = arccot (- 5) + πn

以小数形式表示解

θ = 0.24497 \dots + πn, θ = 2.94419 \dots + πn

作图

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