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受欢迎的三角函数 >

3tan^4(θ)-10tan^2(θ)+3=0

解答

3 tan^{4} (θ) - 10 tan^{2} (θ) + 3 = 0

解答

θ = \frac{π}{3} + πn, θ = \frac{2 π}{3} + πn, θ = 0.52359 \dots + πn, θ = - 0.52359 \dots + πn

+1

度数

θ = 6 0^{\circ} + 18 0^{\circ} n, θ = 12 0^{\circ} + 18 0^{\circ} n, θ = 3 0^{\circ} + 18 0^{\circ} n, θ = - 3 0^{\circ} + 18 0^{\circ} n

求解步骤

3 tan^{4} (θ) - 10 tan^{2} (θ) + 3 = 0

用替代法求解

3 tan^{4} (θ) - 10 tan^{2} (θ) + 3 = 0

令

: tan (θ) = u

3 u^{4} - 10 u^{2} + 3 = 0

3 u^{4} - 10 u^{2} + 3 = 0 : u = 3, u = - 3, u = \frac{1}{3}, u = - \frac{1}{3}

3 u^{4} - 10 u^{2} + 3 = 0

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

3 v^{2} - 10 v + 3 = 0

解

3 v^{2} - 10 v + 3 = 0 : v = 3, v = \frac{1}{3}

3 v^{2} - 10 v + 3 = 0

使用求根公式求解

3 v^{2} - 10 v + 3 = 0

二次方程求根公式:

若

a = 3, b = - 10, c = 3

v_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 3 \cdot 3}{2 \cdot 3}

v_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 3 \cdot 3}{2 \cdot 3}

(- 10)^{2} - 4 \cdot 3 \cdot 3 = 8

(- 10)^{2} - 4 \cdot 3 \cdot 3

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 3 \cdot 3

数字相乘:

4 \cdot 3 \cdot 3 = 36

= 1 0^{2} - 36

1 0^{2} = 100

= 100 - 36

数字相减:

100 - 36 = 64

= 64

因式分解数字:

64 = 8^{2}

= 8^{2}

使用根式运算法则:

n a^{n} = a

8^{2} = 8

= 8

v_{1, 2} = \frac{- ( - 10 ) \pm 8}{2 \cdot 3}

将解分隔开

v_{1} = \frac{- ( - 10 ) + 8}{2 \cdot 3}, v_{2} = \frac{- ( - 10 ) - 8}{2 \cdot 3}

v = \frac{- ( - 10 ) + 8}{2 \cdot 3} : 3

\frac{- ( - 10 ) + 8}{2 \cdot 3}

使用法则

- (- a) = a

= \frac{10 + 8}{2 \cdot 3}

数字相加:

10 + 8 = 18

= \frac{18}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{18}{6}

数字相除:

\frac{18}{6} = 3

= 3

v = \frac{- ( - 10 ) - 8}{2 \cdot 3} : \frac{1}{3}

\frac{- ( - 10 ) - 8}{2 \cdot 3}

使用法则

- (- a) = a

= \frac{10 - 8}{2 \cdot 3}

数字相减:

10 - 8 = 2

= \frac{2}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{2}{6}

约分:

2

= \frac{1}{3}

二次方程组的解是:

v = 3, v = \frac{1}{3}

v = 3, v = \frac{1}{3}

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{1}{3} : u = \frac{1}{3}, u = - \frac{1}{3}

u^{2} = \frac{1}{3}

对于

x^{2} = f (a)

解为

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

u = \frac{1}{3}, u = - \frac{1}{3}

解为

u = 3, u = - 3, u = \frac{1}{3}, u = - \frac{1}{3}

u = tan (θ)

代回

tan (θ) = 3, tan (θ) = - 3, tan (θ) = \frac{1}{3}, tan (θ) = - \frac{1}{3}

tan (θ) = 3, tan (θ) = - 3, tan (θ) = \frac{1}{3}, tan (θ) = - \frac{1}{3}

tan (θ) = 3 : θ = \frac{π}{3} + πn

tan (θ) = 3

tan (θ) = 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}

θ = \frac{π}{3} + πn

θ = \frac{π}{3} + πn

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

tan (θ) = - 3

tan (θ) = - 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}

θ = \frac{2 π}{3} + πn

θ = \frac{2 π}{3} + πn

tan (θ) = \frac{1}{3} : θ = arctan (\frac{1}{3}) + πn

tan (θ) = \frac{1}{3}

使用反三角函数性质

tan (θ) = \frac{1}{3}

tan (θ) = \frac{1}{3}

的通解

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

θ = arctan (\frac{1}{3}) + πn

θ = arctan (\frac{1}{3}) + πn

tan (θ) = - \frac{1}{3} : θ = arctan (- \frac{1}{3}) + πn

tan (θ) = - \frac{1}{3}

使用反三角函数性质

tan (θ) = - \frac{1}{3}

tan (θ) = - \frac{1}{3}

的通解

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

θ = arctan (- \frac{1}{3}) + πn

θ = arctan (- \frac{1}{3}) + πn

合并所有解

θ = \frac{π}{3} + πn, θ = \frac{2 π}{3} + πn, θ = arctan (\frac{1}{3}) + πn, θ = arctan (- \frac{1}{3}) + πn

以小数形式表示解

θ = \frac{π}{3} + πn, θ = \frac{2 π}{3} + πn, θ = 0.52359 \dots + πn, θ = - 0.52359 \dots + πn

作图

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