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受欢迎的三角函数 >

2+3cot^2(θ)-3csc(θ)=5

解答

2 + 3 cot^{2} (θ) - 3 csc (θ) = 5

解答

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn, θ = \frac{3 π}{2} + 2 πn

+1

度数

θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

求解步骤

2 + 3 cot^{2} (θ) - 3 csc (θ) = 5

两边减去

5

3 cot^{2} (θ) - 3 csc (θ) - 3 = 0

使用三角恒等式改写

- 3 + 3 cot^{2} (θ) - 3 csc (θ)

使用毕达哥拉斯恒等式:

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

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

= - 3 + 3 (csc^{2} (θ) - 1) - 3 csc (θ)

化简

- 3 + 3 (csc^{2} (θ) - 1) - 3 csc (θ) : 3 csc^{2} (θ) - 3 csc (θ) - 6

- 3 + 3 (csc^{2} (θ) - 1) - 3 csc (θ)

乘开

3 (csc^{2} (θ) - 1) : 3 csc^{2} (θ) - 3

3 (csc^{2} (θ) - 1)

使用分配律:

a (b - c) = ab - a c

a = 3, b = csc^{2} (θ), c = 1

= 3 csc^{2} (θ) - 3 \cdot 1

数字相乘:

3 \cdot 1 = 3

= 3 csc^{2} (θ) - 3

= - 3 + 3 csc^{2} (θ) - 3 - 3 csc (θ)

化简

- 3 + 3 csc^{2} (θ) - 3 - 3 csc (θ) : 3 csc^{2} (θ) - 3 csc (θ) - 6

- 3 + 3 csc^{2} (θ) - 3 - 3 csc (θ)

对同类项分组

= 3 csc^{2} (θ) - 3 csc (θ) - 3 - 3

数字相减:

- 3 - 3 = - 6

= 3 csc^{2} (θ) - 3 csc (θ) - 6

= 3 csc^{2} (θ) - 3 csc (θ) - 6

= 3 csc^{2} (θ) - 3 csc (θ) - 6

- 6 - 3 csc (θ) + 3 csc^{2} (θ) = 0

用替代法求解

- 6 - 3 csc (θ) + 3 csc^{2} (θ) = 0

令

: csc (θ) = u

- 6 - 3 u + 3 u^{2} = 0

- 6 - 3 u + 3 u^{2} = 0 : u = 2, u = - 1

- 6 - 3 u + 3 u^{2} = 0

改写成标准形式

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

3 u^{2} - 3 u - 6 = 0

使用求根公式求解

3 u^{2} - 3 u - 6 = 0

二次方程求根公式:

若

a = 3, b = - 3, c = - 6

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 3 ( - 6 )}{2 \cdot 3}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 3 ( - 6 )}{2 \cdot 3}

(- 3)^{2} - 4 \cdot 3 (- 6) = 9

(- 3)^{2} - 4 \cdot 3 (- 6)

使用法则

- (- a) = a

= (- 3)^{2} + 4 \cdot 3 \cdot 6

使用指数法则:

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

若

n

是偶数

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

= 3^{2} + 4 \cdot 3 \cdot 6

数字相乘:

4 \cdot 3 \cdot 6 = 72

= 3^{2} + 72

3^{2} = 9

= 9 + 72

数字相加:

9 + 72 = 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{- ( - 3 ) \pm 9}{2 \cdot 3}

将解分隔开

u_{1} = \frac{- ( - 3 ) + 9}{2 \cdot 3}, u_{2} = \frac{- ( - 3 ) - 9}{2 \cdot 3}

u = \frac{- ( - 3 ) + 9}{2 \cdot 3} : 2

\frac{- ( - 3 ) + 9}{2 \cdot 3}

使用法则

- (- a) = a

= \frac{3 + 9}{2 \cdot 3}

数字相加:

3 + 9 = 12

= \frac{12}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{12}{6}

数字相除:

\frac{12}{6} = 2

= 2

u = \frac{- ( - 3 ) - 9}{2 \cdot 3} : - 1

\frac{- ( - 3 ) - 9}{2 \cdot 3}

使用法则

- (- a) = a

= \frac{3 - 9}{2 \cdot 3}

数字相减:

3 - 9 = - 6

= \frac{- 6}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{- 6}{6}

使用分式法则:

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

= - \frac{6}{6}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

u = 2, u = - 1

u = csc (θ)

代回

csc (θ) = 2, csc (θ) = - 1

csc (θ) = 2, csc (θ) = - 1

csc (θ) = 2 : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

csc (θ) = 2

csc (θ) = 2

的通解

csc (x)

周期表（周期为

2 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} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

csc (θ) = - 1 : θ = \frac{3 π}{2} + 2 πn

csc (θ) = - 1

csc (θ) = - 1

的通解

csc (x)

周期表（周期为

2 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} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

θ = \frac{3 π}{2} + 2 πn

θ = \frac{3 π}{2} + 2 πn

合并所有解

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn, θ = \frac{3 π}{2} + 2 πn

作图

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