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受欢迎的三角函数 >

3tan^2(θ)+1= 2/(tan^2(θ))

解答

3 tan^{2} (θ) + 1 = \frac{2}{tan ^{2} ( θ )}

解答

θ = 0.68471 \dots + πn, θ = - 0.68471 \dots + πn

+1

度数

θ = 39.23152 \dots^{\circ} + 18 0^{\circ} n, θ = - 39.23152 \dots^{\circ} + 18 0^{\circ} n

求解步骤

3 tan^{2} (θ) + 1 = \frac{2}{tan ^{2} ( θ )}

用替代法求解

3 tan^{2} (θ) + 1 = \frac{2}{tan ^{2} ( θ )}

令

: tan (θ) = u

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

3 u^{2} + 1 = \frac{2}{u ^{2}} : u = \frac{2}{3}, u = - \frac{2}{3}, u = i, u = - i

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

在两边乘以

u^{2}

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

在两边乘以

u^{2}

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

化简

3 u^{2} u^{2} : 3 u^{4}

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

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

u^{2} u^{2} = u^{2 + 2}

= 3 u^{2 + 2}

数字相加:

2 + 2 = 4

= 3 u^{4}

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

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

解

3 u^{4} + u^{2} = 2 : u = \frac{2}{3}, u = - \frac{2}{3}, u = i, u = - i

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

将

2

para o lado esquerdo

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

两边减去

2

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

化简

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

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

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

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

解

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = 3, b = 1, c = - 2

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

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

1^{2} - 4 \cdot 3 (- 2) = 5

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 3 (- 2)

使用法则

- (- a) = a

= 1 + 4 \cdot 3 \cdot 2

数字相乘:

4 \cdot 3 \cdot 2 = 24

= 1 + 24

数字相加:

1 + 24 = 25

= 25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

5^{2} = 5

= 5

v_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 3}

将解分隔开

v_{1} = \frac{- 1 + 5}{2 \cdot 3}, v_{2} = \frac{- 1 - 5}{2 \cdot 3}

v = \frac{- 1 + 5}{2 \cdot 3} : \frac{2}{3}

\frac{- 1 + 5}{2 \cdot 3}

数字相加/相减:

- 1 + 5 = 4

= \frac{4}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{4}{6}

约分:

2

= \frac{2}{3}

v = \frac{- 1 - 5}{2 \cdot 3} : - 1

\frac{- 1 - 5}{2 \cdot 3}

数字相减:

- 1 - 5 = - 6

= \frac{- 6}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{- 6}{6}

使用分式法则:

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

= - \frac{6}{6}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

v = \frac{2}{3}, v = - 1

v = \frac{2}{3}, v = - 1

代回

v = u^{2}

，求解

u

解

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

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

对于

x^{2} = f (a)

解为

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

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

解

u^{2} = - 1 : u = i, u = - i

u^{2} = - 1

对于

x^{2} = f (a)

解为

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

u = - 1, u = - - 1

化简

- 1 : i

- 1

使用虚数运算法则:

- 1 = i

= i

化简

- - 1 : - i

- - 1

使用虚数运算法则:

- 1 = i

= - i

u = i, u = - i

解为

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

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

验证解

找到无定义的点（奇点）:

u = 0

取

\frac{2}{u ^{2}}

的分母，令其等于零

解

u^{2} = 0 : u = 0

u^{2} = 0

使用法则

x^{n} = 0 \Rightarrow x = 0

u = 0

以下点无定义

u = 0

将不在定义域的点与解相综合:

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

u = tan (θ)

代回

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

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

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

tan (θ) = \frac{2}{3}

使用反三角函数性质

tan (θ) = \frac{2}{3}

tan (θ) = \frac{2}{3}

的通解

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

θ = arctan (\frac{2}{3}) + πn

θ = arctan (\frac{2}{3}) + πn

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

tan (θ) = - \frac{2}{3}

使用反三角函数性质

tan (θ) = - \frac{2}{3}

tan (θ) = - \frac{2}{3}

的通解

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

θ = arctan (- \frac{2}{3}) + πn

θ = arctan (- \frac{2}{3}) + πn

tan (θ) = i :

无解

tan (θ) = i

无解

tan (θ) = - i :

无解

tan (θ) = - i

无解

合并所有解

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

以小数形式表示解

θ = 0.68471 \dots + πn, θ = - 0.68471 \dots + πn

作图

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