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受欢迎的三角函数 >

2tan^4(x)-tan^2(x)-15=0

解答

2 tan^{4} (x) - tan^{2} (x) - 15 = 0

解答

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

+1

度数

x = 6 0^{\circ} + 18 0^{\circ} n, x = 12 0^{\circ} + 18 0^{\circ} n

求解步骤

2 tan^{4} (x) - tan^{2} (x) - 15 = 0

用替代法求解

2 tan^{4} (x) - tan^{2} (x) - 15 = 0

令

: tan (x) = u

2 u^{4} - u^{2} - 15 = 0

2 u^{4} - u^{2} - 15 = 0 : u = 3, u = - 3, u = i \frac{5}{2}, u = - i \frac{5}{2}

2 u^{4} - u^{2} - 15 = 0

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

2 v^{2} - v - 15 = 0

解

2 v^{2} - v - 15 = 0 : v = 3, v = - \frac{5}{2}

2 v^{2} - v - 15 = 0

使用求根公式求解

2 v^{2} - v - 15 = 0

二次方程求根公式:

若

a = 2, b = - 1, c = - 15

v_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 2 ( - 15 )}{2 \cdot 2}

v_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 2 ( - 15 )}{2 \cdot 2}

(- 1)^{2} - 4 \cdot 2 (- 15) = 11

(- 1)^{2} - 4 \cdot 2 (- 15)

使用法则

- (- a) = a

= (- 1)^{2} + 4 \cdot 2 \cdot 15

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

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

若

n

是偶数

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

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 2 \cdot 15 = 120

4 \cdot 2 \cdot 15

数字相乘:

4 \cdot 2 \cdot 15 = 120

= 120

= 1 + 120

数字相加:

1 + 120 = 121

= 121

因式分解数字:

121 = 1 1^{2}

= 1 1^{2}

使用根式运算法则:

n a^{n} = a

1 1^{2} = 11

= 11

v_{1, 2} = \frac{- ( - 1 ) \pm 11}{2 \cdot 2}

将解分隔开

v_{1} = \frac{- ( - 1 ) + 11}{2 \cdot 2}, v_{2} = \frac{- ( - 1 ) - 11}{2 \cdot 2}

v = \frac{- ( - 1 ) + 11}{2 \cdot 2} : 3

\frac{- ( - 1 ) + 11}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{1 + 11}{2 \cdot 2}

数字相加:

1 + 11 = 12

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

数字相乘:

2 \cdot 2 = 4

= \frac{12}{4}

数字相除:

\frac{12}{4} = 3

= 3

v = \frac{- ( - 1 ) - 11}{2 \cdot 2} : - \frac{5}{2}

\frac{- ( - 1 ) - 11}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{1 - 11}{2 \cdot 2}

数字相减:

1 - 11 = - 10

= \frac{- 10}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 10}{4}

使用分式法则:

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

= - \frac{10}{4}

约分:

2

= - \frac{5}{2}

二次方程组的解是:

v = 3, v = - \frac{5}{2}

v = 3, v = - \frac{5}{2}

代回

v = u^{2}

，求解

u

解

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

u^{2} = 3

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = 3, u = - 3

解

u^{2} = - \frac{5}{2} : u = i \frac{5}{2}, u = - i \frac{5}{2}

u^{2} = - \frac{5}{2}

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = - \frac{5}{2}, u = - - \frac{5}{2}

化简

- \frac{5}{2} : i \frac{5}{2}

- \frac{5}{2}

使用根式运算法则:

- a = - 1 a

- \frac{5}{2} = - 1 \frac{5}{2}

= - 1 \frac{5}{2}

使用虚数运算法则:

- 1 = i

= i \frac{5}{2}

化简

- - \frac{5}{2} : - i \frac{5}{2}

- - \frac{5}{2}

化简

- \frac{5}{2} : i \frac{5}{2}

- \frac{5}{2}

使用根式运算法则:

- a = - 1 a

- \frac{5}{2} = - 1 \frac{5}{2}

= - 1 \frac{5}{2}

使用虚数运算法则:

- 1 = i

= i \frac{5}{2}

= - i \frac{5}{2}

u = i \frac{5}{2}, u = - i \frac{5}{2}

解为

u = 3, u = - 3, u = i \frac{5}{2}, u = - i \frac{5}{2}

u = tan (x)

代回

tan (x) = 3, tan (x) = - 3, tan (x) = i \frac{5}{2}, tan (x) = - i \frac{5}{2}

tan (x) = 3, tan (x) = - 3, tan (x) = i \frac{5}{2}, tan (x) = - i \frac{5}{2}

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) = - 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

tan (x) = i \frac{5}{2} :

无解

tan (x) = i \frac{5}{2}

无解

tan (x) = - i \frac{5}{2} :

无解

tan (x) = - i \frac{5}{2}

无解

合并所有解

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

作图

流行的例子

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cos (x) = - \frac{1}{2}

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