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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}{4} + πn, x = \frac{2 π}{3} + πn

+1

度数

x = 4 5^{\circ} + 18 0^{\circ} n, x = 12 0^{\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 = 1, u = - 3

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)

使用法则

- (- a) = a

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

数字相乘:

4 \cdot 1 = 4

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

化简

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

(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

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:

(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} : 1

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

数字相乘:

2 \cdot 1 = 2

= \frac{- ( 3 - 1 ) + 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, - (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

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{- ( 3 - 1 ) - ( 1 + 3 )}{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, - (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

= - 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}

使用分式法则:

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

= - \frac{2 3}{2}

数字相除:

\frac{2}{2} = 1

= - 3

二次方程组的解是:

u = 1, u = - 3

u = tan (x)

代回

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

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

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

tan (x) = - 3 : x = \frac{2 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{2 π}{3} + πn

x = \frac{2 π}{3} + πn

合并所有解

x = \frac{π}{4} + πn, x = \frac{2 π}{3} + πn

流行的例子

sin (θ) = \frac{2}{7}

2 sec (x) - 5 = 0

tan(a)+cot(a)=sec(a)cos(a)

tan (a) + cot (a) = sec (a) cos (a)

-5tan^2(θ)+3=8tan(θ)+6

- 5 tan^{2} (θ) + 3 = 8 tan (θ) + 6

cos(θ)=(-3)/(sqrt(13))

cos (θ) = \frac{- 3}{13}