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受欢迎的三角函数 >

csc^2(θ)-9csc(θ)+20=0

解答

csc^{2} (θ) - 9 csc (θ) + 20 = 0

解答

θ = 0.20135 \dots + 2 πn, θ = π - 0.20135 \dots + 2 πn, θ = 0.25268 \dots + 2 πn, θ = π - 0.25268 \dots + 2 πn

+1

度数

θ = 11.53695 \dots^{\circ} + 36 0^{\circ} n, θ = 168.46304 \dots^{\circ} + 36 0^{\circ} n, θ = 14.47751 \dots^{\circ} + 36 0^{\circ} n, θ = 165.52248 \dots^{\circ} + 36 0^{\circ} n

求解步骤

csc^{2} (θ) - 9 csc (θ) + 20 = 0

用替代法求解

csc^{2} (θ) - 9 csc (θ) + 20 = 0

令

: csc (θ) = u

u^{2} - 9 u + 20 = 0

u^{2} - 9 u + 20 = 0 : u = 5, u = 4

u^{2} - 9 u + 20 = 0

使用求根公式求解

u^{2} - 9 u + 20 = 0

二次方程求根公式:

若

a = 1, b = - 9, c = 20

u_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 1 \cdot 20}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 1 \cdot 20}{2 \cdot 1}

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

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

使用指数法则:

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

若

n

是偶数

(- 9)^{2} = 9^{2}

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

数字相乘:

4 \cdot 1 \cdot 20 = 80

= 9^{2} - 80

9^{2} = 81

= 81 - 80

数字相减:

81 - 80 = 1

= 1

使用法则

1 = 1

= 1

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

将解分隔开

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

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

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

使用法则

- (- a) = a

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

数字相加:

9 + 1 = 10

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

数字相乘:

2 \cdot 1 = 2

= \frac{10}{2}

数字相除:

\frac{10}{2} = 5

= 5

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

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

使用法则

- (- a) = a

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

数字相减:

9 - 1 = 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

二次方程组的解是:

u = 5, u = 4

u = csc (θ)

代回

csc (θ) = 5, csc (θ) = 4

csc (θ) = 5, csc (θ) = 4

csc (θ) = 5 : θ = arccsc (5) + 2 πn, θ = π - arccsc (5) + 2 πn

csc (θ) = 5

使用反三角函数性质

csc (θ) = 5

csc (θ) = 5

的通解

csc (x) = a \Rightarrow x = arccsc (a) + 2 πn, x = π - arccsc (a) + 2 πn

θ = arccsc (5) + 2 πn, θ = π - arccsc (5) + 2 πn

θ = arccsc (5) + 2 πn, θ = π - arccsc (5) + 2 πn

csc (θ) = 4 : θ = arccsc (4) + 2 πn, θ = π - arccsc (4) + 2 πn

csc (θ) = 4

使用反三角函数性质

csc (θ) = 4

csc (θ) = 4

的通解

csc (x) = a \Rightarrow x = arccsc (a) + 2 πn, x = π - arccsc (a) + 2 πn

θ = arccsc (4) + 2 πn, θ = π - arccsc (4) + 2 πn

θ = arccsc (4) + 2 πn, θ = π - arccsc (4) + 2 πn

合并所有解

θ = arccsc (5) + 2 πn, θ = π - arccsc (5) + 2 πn, θ = arccsc (4) + 2 πn, θ = π - arccsc (4) + 2 πn

以小数形式表示解

θ = 0.20135 \dots + 2 πn, θ = π - 0.20135 \dots + 2 πn, θ = 0.25268 \dots + 2 πn, θ = π - 0.25268 \dots + 2 πn

作图

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