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受欢迎的三角函数 >

tan(θ)+1=sqrt(3)+sqrt(3)cot(θ)

解答

tan (θ) + 1 = 3 + 3 cot (θ)

解答

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

+1

度数

θ = 13 5^{\circ} + 18 0^{\circ} n, θ = 6 0^{\circ} + 18 0^{\circ} n

求解步骤

tan (θ) + 1 = 3 + 3 cot (θ)

两边减去

3 + 3 cot (θ)

tan (θ) + 1 - 3 - 3 cot (θ) = 0

使用三角恒等式改写

1 - 3 + tan (θ) - cot (θ) 3

使用基本三角恒等式:

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

= 1 - 3 + \frac{1}{cot ( θ )} - cot (θ) 3

1 + \frac{1}{cot ( θ )} - 3 - cot (θ) 3 = 0

用替代法求解

1 + \frac{1}{cot ( θ )} - 3 - cot (θ) 3 = 0

令

: cot (θ) = u

1 + \frac{1}{u} - 3 - u 3 = 0

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

1 + \frac{1}{u} - 3 - u 3 = 0

在两边乘以

u

1 + \frac{1}{u} - 3 - u 3 = 0

在两边乘以

u

1 \cdot u + \frac{1}{u} u - 3 u - u 3 u = 0 \cdot u

化简

1 \cdot u + \frac{1}{u} u - 3 u - u 3 u = 0 \cdot u

化简

1 \cdot u : u

1 \cdot u

乘以:

1 \cdot u = u

= 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

- u 3 u : - 3 u^{2}

- u 3 u

使用指数法则:

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

uu = u^{1 + 1}

= - 3 u^{1 + 1}

数字相加:

1 + 1 = 2

= - 3 u^{2}

化简

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

u + 1 - 3 u - 3 u^{2} = 0

u + 1 - 3 u - 3 u^{2} = 0

u + 1 - 3 u - 3 u^{2} = 0

解

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

u + 1 - 3 u - 3 u^{2} = 0

改写成标准形式

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

- 3 u^{2} + (1 - 3) u + 1 = 0

使用求根公式求解

- 3 u^{2} + (1 - 3) u + 1 = 0

二次方程求根公式:

若

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

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

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

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

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

使用法则

- (- a) = a

= (1 - 3)^{2} + 43 \cdot 1

数字相乘:

4 \cdot 1 = 4

= (1 - 3)^{2} + 43

乘开

(1 - 3)^{2} + 43 : 4 + 23

(1 - 3)^{2} + 43

(1 - 3)^{2} : 4 - 23

使用完全平方公式:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = 1, b = 3

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

化简

1^{2} - 2 \cdot 1 \cdot 3 + (3)^{2} : 4 - 23

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

使用法则

1^{a} = 1

1^{2} = 1

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

2 \cdot 1 \cdot 3 = 23

2 \cdot 1 \cdot 3

数字相乘:

2 \cdot 1 = 2

= 23

(3)^{2} = 3

(3)^{2}

使用根式运算法则:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= 3^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

分式相乘:

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

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

约分:

2

= 1

= 3

= 1 - 23 + 3

数字相加:

1 + 3 = 4

= 4 - 23

= 4 - 23

= 4 - 23 + 43

同类项相加:

- 23 + 43 = 23

= 4 + 23

= 4 + 23

= 3 + 23 + 1

= (3)^{2} + 23 + (1)^{2}

1 = 1

1

使用法则

1 = 1

= 1

= (3)^{2} + 23 + 1^{2}

23 \cdot 1 = 23

23 \cdot 1

数字相乘:

2 \cdot 1 = 2

= 23

= (3)^{2} + 23 \cdot 1 + 1^{2}

使用完全平方公式:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

(3)^{2} + 23 \cdot 1 + 1^{2} = (3 + 1)^{2}

= (3 + 1)^{2}

使用根式运算法则:

n a^{n} = a

(3 + 1)^{2} = 3 + 1

= 3 + 1

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

将解分隔开

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

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

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

去除括号:

(- a) = - a

= \frac{- ( 1 - 3 ) + 3 + 1}{- 2 3}

使用分式法则:

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

= - \frac{- ( 1 - 3 ) + 3 + 1}{2 3}

乘开

- (1 - 3) + 3 + 1 : 23

- (1 - 3) + 3 + 1

- (1 - 3) : - 1 + 3

- (1 - 3)

打开括号

= - (1) - (- 3)

使用加减运算法则

- (- a) = a, - (a) = - a

= - 1 + 3

= - 1 + 3 + 3 + 1

化简

- 1 + 3 + 3 + 1 : 23

- 1 + 3 + 3 + 1

同类项相加:

3 + 3 = 23

= - 1 + 23 + 1

- 1 + 1 = 0

= 23

= 23

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

使用法则

\frac{a}{a} = 1

= - 1

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

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

去除括号:

(- a) = - a

= \frac{- ( 1 - 3 ) - ( 3 + 1 )}{- 2 3}

使用分式法则:

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

- (1 - 3) - (3 + 1) = - ((1 - 3) + (1 + 3))

= \frac{( 1 - 3 ) + ( 1 + 3 )}{2 3}

去除括号:

(a) = a

= \frac{1 - 3 + 1 + 3}{2 3}

1 - 3 + 1 + 3 = 2

1 - 3 + 1 + 3

同类项相加:

- 3 + 3 = 0

= 1 + 1

数字相加:

1 + 1 = 2

= 2

= \frac{2}{2 3}

数字相除:

\frac{2}{2} = 1

= \frac{1}{3}

\frac{1}{3}

有理化:

\frac{3}{3}

\frac{1}{3}

乘以共轭根式

\frac{3}{3}

= \frac{1 \cdot 3}{3 3}

1 \cdot 3 = 3

33 = 3

33

使用根式运算法则:

a a = a

33 = 3

= 3

= \frac{3}{3}

= \frac{3}{3}

二次方程组的解是:

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

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

验证解

找到无定义的点（奇点）:

u = 0

取

1 + \frac{1}{u} - 3 - u 3

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

u = cot (θ)

代回

cot (θ) = - 1, cot (θ) = \frac{3}{3}

cot (θ) = - 1, cot (θ) = \frac{3}{3}

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

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

cot (θ) = \frac{3}{3}

cot (θ) = \frac{3}{3}

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{π}{3} + πn

θ = \frac{π}{3} + πn

合并所有解

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

作图

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