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受欢迎的三角函数 >

tan(θ)-7=-12cot(θ)

解答

tan (θ) - 7 = - 12 cot (θ)

解答

θ = 1.24904 \dots + πn, θ = 1.32581 \dots + πn

+1

度数

θ = 71.56505 \dots^{\circ} + 18 0^{\circ} n, θ = 75.96375 \dots^{\circ} + 18 0^{\circ} n

求解步骤

tan (θ) - 7 = - 12 cot (θ)

两边减去

- 12 cot (θ)

tan (θ) - 7 + 12 cot (θ) = 0

使用三角恒等式改写

- 7 + tan (θ) + 12 cot (θ)

使用基本三角恒等式:

tan (x) = \frac{1}{cot ( x )}

= - 7 + \frac{1}{cot ( θ )} + 12 cot (θ)

- 7 + \frac{1}{cot ( θ )} + 12 cot (θ) = 0

用替代法求解

- 7 + \frac{1}{cot ( θ )} + 12 cot (θ) = 0

令

: cot (θ) = u

- 7 + \frac{1}{u} + 12 u = 0

- 7 + \frac{1}{u} + 12 u = 0 : u = \frac{1}{3}, u = \frac{1}{4}

- 7 + \frac{1}{u} + 12 u = 0

在两边乘以

u

- 7 + \frac{1}{u} + 12 u = 0

在两边乘以

u

- 7 u + \frac{1}{u} u + 12 uu = 0 \cdot u

化简

- 7 u + \frac{1}{u} u + 12 uu = 0 \cdot u

化简

\frac{1}{u} u : 1

\frac{1}{u} u

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot u}{u}

约分:

u

= 1

化简

12 uu : 12 u^{2}

12 uu

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

uu = u^{1 + 1}

= 12 u^{1 + 1}

数字相加:

1 + 1 = 2

= 12 u^{2}

化简

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

- 7 u + 1 + 12 u^{2} = 0

- 7 u + 1 + 12 u^{2} = 0

- 7 u + 1 + 12 u^{2} = 0

解

- 7 u + 1 + 12 u^{2} = 0 : u = \frac{1}{3}, u = \frac{1}{4}

- 7 u + 1 + 12 u^{2} = 0

改写成标准形式

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

12 u^{2} - 7 u + 1 = 0

使用求根公式求解

12 u^{2} - 7 u + 1 = 0

二次方程求根公式:

若

a = 12, b = - 7, c = 1

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 12 \cdot 1}{2 \cdot 12}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 12 \cdot 1}{2 \cdot 12}

(- 7)^{2} - 4 \cdot 12 \cdot 1 = 1

(- 7)^{2} - 4 \cdot 12 \cdot 1

使用指数法则:

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

若

n

是偶数

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

= 7^{2} - 4 \cdot 12 \cdot 1

数字相乘:

4 \cdot 12 \cdot 1 = 48

= 7^{2} - 48

7^{2} = 49

= 49 - 48

数字相减:

49 - 48 = 1

= 1

使用法则

1 = 1

= 1

u_{1, 2} = \frac{- ( - 7 ) \pm 1}{2 \cdot 12}

将解分隔开

u_{1} = \frac{- ( - 7 ) + 1}{2 \cdot 12}, u_{2} = \frac{- ( - 7 ) - 1}{2 \cdot 12}

u = \frac{- ( - 7 ) + 1}{2 \cdot 12} : \frac{1}{3}

\frac{- ( - 7 ) + 1}{2 \cdot 12}

使用法则

- (- a) = a

= \frac{7 + 1}{2 \cdot 12}

数字相加:

7 + 1 = 8

= \frac{8}{2 \cdot 12}

数字相乘:

2 \cdot 12 = 24

= \frac{8}{24}

约分:

8

= \frac{1}{3}

u = \frac{- ( - 7 ) - 1}{2 \cdot 12} : \frac{1}{4}

\frac{- ( - 7 ) - 1}{2 \cdot 12}

使用法则

- (- a) = a

= \frac{7 - 1}{2 \cdot 12}

数字相减:

7 - 1 = 6

= \frac{6}{2 \cdot 12}

数字相乘:

2 \cdot 12 = 24

= \frac{6}{24}

约分:

6

= \frac{1}{4}

二次方程组的解是:

u = \frac{1}{3}, u = \frac{1}{4}

u = \frac{1}{3}, u = \frac{1}{4}

验证解

找到无定义的点（奇点）:

u = 0

取

- 7 + \frac{1}{u} + 12 u

的分母，令其等于零

u = 0

以下点无定义

u = 0

将不在定义域的点与解相综合:

u = \frac{1}{3}, u = \frac{1}{4}

u = cot (θ)

代回

cot (θ) = \frac{1}{3}, cot (θ) = \frac{1}{4}

cot (θ) = \frac{1}{3}, cot (θ) = \frac{1}{4}

cot (θ) = \frac{1}{3} : θ = arccot (\frac{1}{3}) + πn

cot (θ) = \frac{1}{3}

使用反三角函数性质

cot (θ) = \frac{1}{3}

cot (θ) = \frac{1}{3}

的通解

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

θ = arccot (\frac{1}{3}) + πn

θ = arccot (\frac{1}{3}) + πn

cot (θ) = \frac{1}{4} : θ = arccot (\frac{1}{4}) + πn

cot (θ) = \frac{1}{4}

使用反三角函数性质

cot (θ) = \frac{1}{4}

cot (θ) = \frac{1}{4}

的通解

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

θ = arccot (\frac{1}{4}) + πn

θ = arccot (\frac{1}{4}) + πn

合并所有解

θ = arccot (\frac{1}{3}) + πn, θ = arccot (\frac{1}{4}) + πn

以小数形式表示解

θ = 1.24904 \dots + πn, θ = 1.32581 \dots + πn

作图

流行的例子

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6 sin (b) + 12 = 0

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sin(x)(sin(x)-1)=3cos^2(x)

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