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受欢迎的三角函数 >

tan^4(x)+1=4tan^2(x)-2

解答

tan^{4} (x) + 1 = 4 tan^{2} (x) - 2

解答

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

+1

度数

x = 6 0^{\circ} + 18 0^{\circ} n, x = 12 0^{\circ} + 18 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n

求解步骤

tan^{4} (x) + 1 = 4 tan^{2} (x) - 2

用替代法求解

tan^{4} (x) + 1 = 4 tan^{2} (x) - 2

令

: tan (x) = u

u^{4} + 1 = 4 u^{2} - 2

u^{4} + 1 = 4 u^{2} - 2 : u = 3, u = - 3, u = 1, u = - 1

u^{4} + 1 = 4 u^{2} - 2

将

2

para o lado esquerdo

u^{4} + 1 = 4 u^{2} - 2

两边加上

2

u^{4} + 1 + 2 = 4 u^{2} - 2 + 2

化简

u^{4} + 3 = 4 u^{2}

u^{4} + 3 = 4 u^{2}

将

4 u^{2}

para o lado esquerdo

u^{4} + 3 = 4 u^{2}

两边减去

4 u^{2}

u^{4} + 3 - 4 u^{2} = 4 u^{2} - 4 u^{2}

化简

u^{4} + 3 - 4 u^{2} = 0

u^{4} + 3 - 4 u^{2} = 0

改写成标准形式

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

u^{4} - 4 u^{2} + 3 = 0

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

v^{2} - 4 v + 3 = 0

解

v^{2} - 4 v + 3 = 0 : v = 3, v = 1

v^{2} - 4 v + 3 = 0

使用求根公式求解

v^{2} - 4 v + 3 = 0

二次方程求根公式:

若

a = 1, b = - 4, c = 3

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

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

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

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

使用指数法则:

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

若

n

是偶数

(- 4)^{2} = 4^{2}

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

数字相乘:

4 \cdot 1 \cdot 3 = 12

= 4^{2} - 12

4^{2} = 16

= 16 - 12

数字相减:

16 - 12 = 4

= 4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

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

将解分隔开

v_{1} = \frac{- ( - 4 ) + 2}{2 \cdot 1}, v_{2} = \frac{- ( - 4 ) - 2}{2 \cdot 1}

v = \frac{- ( - 4 ) + 2}{2 \cdot 1} : 3

\frac{- ( - 4 ) + 2}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{4 + 2}{2 \cdot 1}

数字相加:

4 + 2 = 6

= \frac{6}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{6}{2}

数字相除:

\frac{6}{2} = 3

= 3

v = \frac{- ( - 4 ) - 2}{2 \cdot 1} : 1

\frac{- ( - 4 ) - 2}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{4 - 2}{2 \cdot 1}

数字相减:

4 - 2 = 2

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

数字相乘:

2 \cdot 1 = 2

= \frac{2}{2}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

v = 3, v = 1

v = 3, v = 1

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} = 1 : u = 1, u = - 1

u^{2} = 1

对于

x^{2} = f (a)

解为

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

u = 1, u = - 1

1 = 1

1

使用法则

1 = 1

= 1

- 1 = - 1

- 1

使用法则

1 = 1

= - 1

u = 1, u = - 1

解为

u = 3, u = - 3, u = 1, u = - 1

u = tan (x)

代回

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

tan (x) = 3, tan (x) = - 3, tan (x) = 1, 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) = - 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) = 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) = - 1 : x = \frac{3 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{3 π}{4} + πn

x = \frac{3 π}{4} + πn

合并所有解

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

作图

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