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受欢迎的三角函数 >

csc^2(θ)=7cot(θ)+9

解答

csc^{2} (θ) = 7 cot (θ) + 9

解答

θ = 0.12435 \dots + πn, θ = \frac{3 π}{4} + πn

+1

度数

θ = 7.12501 \dots^{\circ} + 18 0^{\circ} n, θ = 13 5^{\circ} + 18 0^{\circ} n

求解步骤

csc^{2} (θ) = 7 cot (θ) + 9

两边减去

7 cot (θ) + 9

csc^{2} (θ) - 7 cot (θ) - 9 = 0

使用三角恒等式改写

- 9 + csc^{2} (θ) - 7 cot (θ)

使用毕达哥拉斯恒等式:

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

= - 9 + 1 + cot^{2} (θ) - 7 cot (θ)

化简

= cot^{2} (θ) - 7 cot (θ) - 8

- 8 + cot^{2} (θ) - 7 cot (θ) = 0

用替代法求解

- 8 + cot^{2} (θ) - 7 cot (θ) = 0

令

: cot (θ) = u

- 8 + u^{2} - 7 u = 0

- 8 + u^{2} - 7 u = 0 : u = 8, u = - 1

- 8 + u^{2} - 7 u = 0

改写成标准形式

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

u^{2} - 7 u - 8 = 0

使用求根公式求解

u^{2} - 7 u - 8 = 0

二次方程求根公式:

若

a = 1, b = - 7, c = - 8

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

(- 7)^{2} - 4 \cdot 1 \cdot (- 8) = 9

(- 7)^{2} - 4 \cdot 1 \cdot (- 8)

使用法则

- (- a) = a

= (- 7)^{2} + 4 \cdot 1 \cdot 8

使用指数法则:

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

若

n

是偶数

(- 7)^{2} = 7^{2}

= 7^{2} + 4 \cdot 1 \cdot 8

数字相乘:

4 \cdot 1 \cdot 8 = 32

= 7^{2} + 32

7^{2} = 49

= 49 + 32

数字相加:

49 + 32 = 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{- ( - 7 ) \pm 9}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- ( - 7 ) + 9}{2 \cdot 1}, u_{2} = \frac{- ( - 7 ) - 9}{2 \cdot 1}

u = \frac{- ( - 7 ) + 9}{2 \cdot 1} : 8

\frac{- ( - 7 ) + 9}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{7 + 9}{2 \cdot 1}

数字相加:

7 + 9 = 16

= \frac{16}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{16}{2}

数字相除:

\frac{16}{2} = 8

= 8

u = \frac{- ( - 7 ) - 9}{2 \cdot 1} : - 1

\frac{- ( - 7 ) - 9}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{7 - 9}{2 \cdot 1}

数字相减:

7 - 9 = - 2

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

数字相乘:

2 \cdot 1 = 2

= \frac{- 2}{2}

使用分式法则:

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

= - \frac{2}{2}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

u = 8, u = - 1

u = cot (θ)

代回

cot (θ) = 8, cot (θ) = - 1

cot (θ) = 8, cot (θ) = - 1

cot (θ) = 8 : θ = arccot (8) + πn

cot (θ) = 8

使用反三角函数性质

cot (θ) = 8

cot (θ) = 8

的通解

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

θ = arccot (8) + πn

θ = arccot (8) + πn

cot (θ) = - 1 : θ = \frac{3 π}{4} + πn

cot (θ) = - 1

cot (θ) = - 1

的通解

cot (x)

周期表（周期为

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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

合并所有解

θ = arccot (8) + πn, θ = \frac{3 π}{4} + πn

以小数形式表示解

θ = 0.12435 \dots + πn, θ = \frac{3 π}{4} + πn

作图

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