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受欢迎的三角函数 >

tan^2(x)-(sqrt(3)+1)tan(x)+sqrt(3)=0

解答

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

解答

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

+1

度数

x = 6 0^{\circ} + 18 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

求解步骤

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

用替代法求解

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

令

: tan (x) = u

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

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

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

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

数字相乘:

4 \cdot 1 = 4

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

乘开

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

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

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

使用完全平方公式:

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

a = - 3, b = 1

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

化简

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

使用法则

- (- a) = a

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

(- 3)^{2} = 3

(- 3)^{2}

使用指数法则:

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

若

n

是偶数

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

= (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

23 \cdot 1 = 23

23 \cdot 1

数字相乘:

2 \cdot 1 = 2

= 23

= 3 + 23 + 1

数字相加:

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

将解分隔开

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

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

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

数字相乘:

2 \cdot 1 = 2

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

乘开

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

- (- 3 - 1) + 3 - 1

- (- 3 - 1) : 3 + 1

- (- 3 - 1)

打开括号

= - (- 3) - (- 1)

使用加减运算法则

- (- a) = a

= 3 + 1

= 3 + 1 + 3 - 1

化简

3 + 1 + 3 - 1 : 23

3 + 1 + 3 - 1

同类项相加:

3 + 3 = 23

= 23 + 1 - 1

1 - 1 = 0

= 23

= 23

= \frac{2 3}{2}

数字相除:

\frac{2}{2} = 1

= 3

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

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

数字相乘:

2 \cdot 1 = 2

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

乘开

- (- 3 - 1) - (3 - 1) : 2

- (- 3 - 1) - (3 - 1)

- (- 3 - 1) : 3 + 1

- (- 3 - 1)

打开括号

= - (- 3) - (- 1)

使用加减运算法则

- (- a) = a

= 3 + 1

= 3 + 1 - (3 - 1)

- (3 - 1) : - 3 + 1

- (3 - 1)

打开括号

= - (3) - (- 1)

使用加减运算法则

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

= - 3 + 1

= 3 + 1 - 3 + 1

化简

3 + 1 - 3 + 1 : 2

3 + 1 - 3 + 1

同类项相加:

3 - 3 = 0

= 1 + 1

数字相加:

1 + 1 = 2

= 2

= 2

= \frac{2}{2}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

u = 3, u = 1

u = tan (x)

代回

tan (x) = 3, tan (x) = 1

tan (x) = 3, tan (x) = 1

tan (x) = 3 : x = \frac{π}{3} + πn

tan (x) = 3

tan (x) = 3

的通解

tan (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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{3} + πn

x = \frac{π}{3} + πn

tan (x) = 1 : x = \frac{π}{4} + πn

tan (x) = 1

tan (x) = 1

的通解

tan (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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

合并所有解

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

作图

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