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5tan^4(x)-10tan^2(x)+1=0

解答

5 tan^{4} (x) - 10 tan^{2} (x) + 1 = 0

解答

x = 0.94247 \dots + πn, x = - 0.94247 \dots + πn, x = 0.31415 \dots + πn, x = - 0.31415 \dots + πn

+1

度数

x = 5 4^{\circ} + 18 0^{\circ} n, x = - 5 4^{\circ} + 18 0^{\circ} n, x = 1 8^{\circ} + 18 0^{\circ} n, x = - 1 8^{\circ} + 18 0^{\circ} n

求解步骤

5 tan^{4} (x) - 10 tan^{2} (x) + 1 = 0

用替代法求解

5 tan^{4} (x) - 10 tan^{2} (x) + 1 = 0

令

: tan (x) = u

5 u^{4} - 10 u^{2} + 1 = 0

5 u^{4} - 10 u^{2} + 1 = 0 : u = \frac{5 + 2 5}{5}, u = - \frac{5 + 2 5}{5}, u = \frac{5 - 2 5}{5}, u = - \frac{5 - 2 5}{5}

5 u^{4} - 10 u^{2} + 1 = 0

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

5 v^{2} - 10 v + 1 = 0

解

5 v^{2} - 10 v + 1 = 0 : v = \frac{5 + 2 5}{5}, v = \frac{5 - 2 5}{5}

5 v^{2} - 10 v + 1 = 0

使用求根公式求解

5 v^{2} - 10 v + 1 = 0

二次方程求根公式:

若

a = 5, b = - 10, c = 1

v_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 5 \cdot 1}{2 \cdot 5}

v_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 5 \cdot 1}{2 \cdot 5}

(- 10)^{2} - 4 \cdot 5 \cdot 1 = 45

(- 10)^{2} - 4 \cdot 5 \cdot 1

使用指数法则:

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

若

n

是偶数

(- 10)^{2} = 1 0^{2}

= 1 0^{2} - 4 \cdot 5 \cdot 1

数字相乘:

4 \cdot 5 \cdot 1 = 20

= 1 0^{2} - 20

1 0^{2} = 100

= 100 - 20

数字相减:

100 - 20 = 80

= 80

80

质因数分解:

2^{4} \cdot 5

80

80

除以

280 = 40 \cdot 2

= 2 \cdot 40

40

除以

240 = 20 \cdot 2

= 2 \cdot 2 \cdot 20

20

除以

220 = 10 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 10

10

除以

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 5

2, 5

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 5

= 2^{4} \cdot 5

= 2^{4} \cdot 5

使用根式运算法则:

= 5 2^{4}

使用根式运算法则:

2^{4} = 2^{\frac{4}{2}} = 2^{2}

= 2^{2} 5

整理后得

= 45

v_{1, 2} = \frac{- ( - 10 ) \pm 4 5}{2 \cdot 5}

将解分隔开

v_{1} = \frac{- ( - 10 ) + 4 5}{2 \cdot 5}, v_{2} = \frac{- ( - 10 ) - 4 5}{2 \cdot 5}

v = \frac{- ( - 10 ) + 4 5}{2 \cdot 5} : \frac{5 + 2 5}{5}

\frac{- ( - 10 ) + 4 5}{2 \cdot 5}

使用法则

- (- a) = a

= \frac{10 + 4 5}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{10 + 4 5}{10}

分解

10 + 45 : 2 (5 + 25)

10 + 45

改写为

= 2 \cdot 5 + 2 \cdot 25

因式分解出通项

2

= 2 (5 + 25)

= \frac{2 ( 5 + 2 5 )}{10}

约分:

2

= \frac{5 + 2 5}{5}

v = \frac{- ( - 10 ) - 4 5}{2 \cdot 5} : \frac{5 - 2 5}{5}

\frac{- ( - 10 ) - 4 5}{2 \cdot 5}

使用法则

- (- a) = a

= \frac{10 - 4 5}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{10 - 4 5}{10}

分解

10 - 45 : 2 (5 - 25)

10 - 45

改写为

= 2 \cdot 5 - 2 \cdot 25

因式分解出通项

2

= 2 (5 - 25)

= \frac{2 ( 5 - 2 5 )}{10}

约分:

2

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

二次方程组的解是:

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

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

代回

v = u^{2}

，求解

u

解

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

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

对于

x^{2} = f (a)

解为

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

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

解

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

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

对于

x^{2} = f (a)

解为

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

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

解为

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

u = tan (x)

代回

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

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

tan (x) = \frac{5 + 2 5}{5} : x = arctan \frac{5 + 2 5}{5} + πn

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

使用反三角函数性质

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

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

的通解

tan (x) = a \Rightarrow x = arctan (a) + πn

x = arctan \frac{5 + 2 5}{5} + πn

x = arctan \frac{5 + 2 5}{5} + πn

tan (x) = - \frac{5 + 2 5}{5} : x = arctan - \frac{5 + 2 5}{5} + πn

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

使用反三角函数性质

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

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

的通解

tan (x) = - a \Rightarrow x = arctan (- a) + πn

x = arctan - \frac{5 + 2 5}{5} + πn

x = arctan - \frac{5 + 2 5}{5} + πn

tan (x) = \frac{5 - 2 5}{5} : x = arctan \frac{5 - 2 5}{5} + πn

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

使用反三角函数性质

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

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

的通解

tan (x) = a \Rightarrow x = arctan (a) + πn

x = arctan \frac{5 - 2 5}{5} + πn

x = arctan \frac{5 - 2 5}{5} + πn

tan (x) = - \frac{5 - 2 5}{5} : x = arctan - \frac{5 - 2 5}{5} + πn

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

使用反三角函数性质

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

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

的通解

tan (x) = - a \Rightarrow x = arctan (- a) + πn

x = arctan - \frac{5 - 2 5}{5} + πn

x = arctan - \frac{5 - 2 5}{5} + πn

合并所有解

x = arctan \frac{5 + 2 5}{5} + πn, x = arctan - \frac{5 + 2 5}{5} + πn, x = arctan \frac{5 - 2 5}{5} + πn, x = arctan - \frac{5 - 2 5}{5} + πn

以小数形式表示解

x = 0.94247 \dots + πn, x = - 0.94247 \dots + πn, x = 0.31415 \dots + πn, x = - 0.31415 \dots + πn

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