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受欢迎的三角函数 >

tan^4(θ)-20tan^2(θ)+64=0

解答

tan^{4} (θ) - 20 tan^{2} (θ) + 64 = 0

解答

θ = 1.32581 \dots + πn, θ = - 1.32581 \dots + πn, θ = 1.10714 \dots + πn, θ = - 1.10714 \dots + πn

+1

度数

θ = 75.96375 \dots^{\circ} + 18 0^{\circ} n, θ = - 75.96375 \dots^{\circ} + 18 0^{\circ} n, θ = 63.43494 \dots^{\circ} + 18 0^{\circ} n, θ = - 63.43494 \dots^{\circ} + 18 0^{\circ} n

求解步骤

tan^{4} (θ) - 20 tan^{2} (θ) + 64 = 0

用替代法求解

tan^{4} (θ) - 20 tan^{2} (θ) + 64 = 0

令

: tan (θ) = u

u^{4} - 20 u^{2} + 64 = 0

u^{4} - 20 u^{2} + 64 = 0 : u = 4, u = - 4, u = 2, u = - 2

u^{4} - 20 u^{2} + 64 = 0

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

v^{2} - 20 v + 64 = 0

解

v^{2} - 20 v + 64 = 0 : v = 16, v = 4

v^{2} - 20 v + 64 = 0

使用求根公式求解

v^{2} - 20 v + 64 = 0

二次方程求根公式:

若

a = 1, b = - 20, c = 64

v_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 1 \cdot 64}{2 \cdot 1}

v_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 1 \cdot 64}{2 \cdot 1}

(- 20)^{2} - 4 \cdot 1 \cdot 64 = 12

(- 20)^{2} - 4 \cdot 1 \cdot 64

使用指数法则:

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

若

n

是偶数

(- 20)^{2} = 2 0^{2}

= 2 0^{2} - 4 \cdot 1 \cdot 64

数字相乘:

4 \cdot 1 \cdot 64 = 256

= 2 0^{2} - 256

2 0^{2} = 400

= 400 - 256

数字相减:

400 - 256 = 144

= 144

因式分解数字:

144 = 1 2^{2}

= 1 2^{2}

使用根式运算法则:

n a^{n} = a

1 2^{2} = 12

= 12

v_{1, 2} = \frac{- ( - 20 ) \pm 12}{2 \cdot 1}

将解分隔开

v_{1} = \frac{- ( - 20 ) + 12}{2 \cdot 1}, v_{2} = \frac{- ( - 20 ) - 12}{2 \cdot 1}

v = \frac{- ( - 20 ) + 12}{2 \cdot 1} : 16

\frac{- ( - 20 ) + 12}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{20 + 12}{2 \cdot 1}

数字相加:

20 + 12 = 32

= \frac{32}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{32}{2}

数字相除:

\frac{32}{2} = 16

= 16

v = \frac{- ( - 20 ) - 12}{2 \cdot 1} : 4

\frac{- ( - 20 ) - 12}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{20 - 12}{2 \cdot 1}

数字相减:

20 - 12 = 8

= \frac{8}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{8}{2}

数字相除:

\frac{8}{2} = 4

= 4

二次方程组的解是:

v = 16, v = 4

v = 16, v = 4

代回

v = u^{2}

，求解

u

解

u^{2} = 16 : u = 4, u = - 4

u^{2} = 16

对于

x^{2} = f (a)

解为

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

u = 16, u = - 16

16 = 4

16

因式分解数字:

16 = 4^{2}

= 4^{2}

使用根式运算法则:

n a^{n} = a

4^{2} = 4

= 4

- 16 = - 4

- 16

16 = 4

16

因式分解数字:

16 = 4^{2}

= 4^{2}

使用根式运算法则:

n a^{n} = a

4^{2} = 4

= 4

= - 4

u = 4, u = - 4

解

u^{2} = 4 : u = 2, u = - 2

u^{2} = 4

对于

x^{2} = f (a)

解为

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

u = 4, u = - 4

4 = 2

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

- 4 = - 2

- 4

4 = 2

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

= - 2

u = 2, u = - 2

解为

u = 4, u = - 4, u = 2, u = - 2

u = tan (θ)

代回

tan (θ) = 4, tan (θ) = - 4, tan (θ) = 2, tan (θ) = - 2

tan (θ) = 4, tan (θ) = - 4, tan (θ) = 2, tan (θ) = - 2

tan (θ) = 4 : θ = arctan (4) + πn

tan (θ) = 4

使用反三角函数性质

tan (θ) = 4

tan (θ) = 4

的通解

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

θ = arctan (4) + πn

θ = arctan (4) + πn

tan (θ) = - 4 : θ = arctan (- 4) + πn

tan (θ) = - 4

使用反三角函数性质

tan (θ) = - 4

tan (θ) = - 4

的通解

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

θ = arctan (- 4) + πn

θ = arctan (- 4) + πn

tan (θ) = 2 : θ = arctan (2) + πn

tan (θ) = 2

使用反三角函数性质

tan (θ) = 2

tan (θ) = 2

的通解

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

θ = arctan (2) + πn

θ = arctan (2) + πn

tan (θ) = - 2 : θ = arctan (- 2) + πn

tan (θ) = - 2

使用反三角函数性质

tan (θ) = - 2

tan (θ) = - 2

的通解

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

θ = arctan (- 2) + πn

θ = arctan (- 2) + πn

合并所有解

θ = arctan (4) + πn, θ = arctan (- 4) + πn, θ = arctan (2) + πn, θ = arctan (- 2) + πn

以小数形式表示解

θ = 1.32581 \dots + πn, θ = - 1.32581 \dots + πn, θ = 1.10714 \dots + πn, θ = - 1.10714 \dots + πn

作图

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