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csc^2(θ)=3cot(θ)+5

解答

csc^{2} (θ) = 3 cot (θ) + 5

解答

θ = 0.24497 \dots + πn, θ = \frac{3 π}{4} + πn

+1

度数

θ = 14.03624 \dots^{\circ} + 18 0^{\circ} n, θ = 13 5^{\circ} + 18 0^{\circ} n

求解步骤

csc^{2} (θ) = 3 cot (θ) + 5

两边减去

3 cot (θ) + 5

csc^{2} (θ) - 3 cot (θ) - 5 = 0

使用三角恒等式改写

- 5 + csc^{2} (θ) - 3 cot (θ)

使用毕达哥拉斯恒等式:

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

= - 5 + 1 + cot^{2} (θ) - 3 cot (θ)

化简

= cot^{2} (θ) - 3 cot (θ) - 4

- 4 + cot^{2} (θ) - 3 cot (θ) = 0

用替代法求解

- 4 + cot^{2} (θ) - 3 cot (θ) = 0

令

: cot (θ) = u

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

- 4 + u^{2} - 3 u = 0 : u = 4, u = - 1

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

改写成标准形式

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

u^{2} - 3 u - 4 = 0

使用求根公式求解

u^{2} - 3 u - 4 = 0

二次方程求根公式:

若

a = 1, b = - 3, c = - 4

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

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

(- 3)^{2} - 4 \cdot 1 \cdot (- 4) = 5

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

使用法则

- (- a) = a

= (- 3)^{2} + 4 \cdot 1 \cdot 4

使用指数法则:

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

若

n

是偶数

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

= 3^{2} + 4 \cdot 1 \cdot 4

数字相乘:

4 \cdot 1 \cdot 4 = 16

= 3^{2} + 16

3^{2} = 9

= 9 + 16

数字相加:

9 + 16 = 25

= 25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

5^{2} = 5

= 5

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

将解分隔开

u_{1} = \frac{- ( - 3 ) + 5}{2 \cdot 1}, u_{2} = \frac{- ( - 3 ) - 5}{2 \cdot 1}

u = \frac{- ( - 3 ) + 5}{2 \cdot 1} : 4

\frac{- ( - 3 ) + 5}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{3 + 5}{2 \cdot 1}

数字相加:

3 + 5 = 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{- ( - 3 ) - 5}{2 \cdot 1} : - 1

\frac{- ( - 3 ) - 5}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{3 - 5}{2 \cdot 1}

数字相减:

3 - 5 = - 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 = 4, u = - 1

u = cot (θ)

代回

cot (θ) = 4, cot (θ) = - 1

cot (θ) = 4, cot (θ) = - 1

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

以小数形式表示解

θ = 0.24497 \dots + πn, θ = \frac{3 π}{4} + πn

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