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受欢迎的三角函数 >

2tan^2(x)-3cot^2(x)=5

解答

2 tan^{2} (x) - 3 cot^{2} (x) = 5

解答

x = 1.04719 \dots + πn, x = 2.09439 \dots + πn

+1

度数

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

求解步骤

2 tan^{2} (x) - 3 cot^{2} (x) = 5

两边减去

5

2 tan^{2} (x) - 3 cot^{2} (x) - 5 = 0

使用三角恒等式改写

- 5 + 2 tan^{2} (x) - 3 cot^{2} (x)

使用基本三角恒等式:

tan (x) = \frac{1}{cot ( x )}

= - 5 + 2 (\frac{1}{cot ( x )})^{2} - 3 cot^{2} (x)

2 (\frac{1}{cot ( x )})^{2} = \frac{2}{cot ^{2} ( x )}

2 (\frac{1}{cot ( x )})^{2}

(\frac{1}{cot ( x )})^{2} = \frac{1}{cot ^{2} ( x )}

(\frac{1}{cot ( x )})^{2}

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{cot ^{2} ( x )}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{cot ^{2} ( x )}

= 2 \cdot \frac{1}{cot ^{2} ( x )}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{cot ^{2} ( x )}

数字相乘:

1 \cdot 2 = 2

= \frac{2}{cot ^{2} ( x )}

= - 5 + \frac{2}{cot ^{2} ( x )} - 3 cot^{2} (x)

- 5 + \frac{2}{cot ^{2} ( x )} - 3 cot^{2} (x) = 0

用替代法求解

- 5 + \frac{2}{cot ^{2} ( x )} - 3 cot^{2} (x) = 0

令

: cot (x) = u

- 5 + \frac{2}{u ^{2}} - 3 u^{2} = 0

- 5 + \frac{2}{u ^{2}} - 3 u^{2} = 0 : u = 2 i, u = - 2 i, u = \frac{1}{3}, u = - \frac{1}{3}

- 5 + \frac{2}{u ^{2}} - 3 u^{2} = 0

在两边乘以

u^{2}

- 5 + \frac{2}{u ^{2}} - 3 u^{2} = 0

在两边乘以

u^{2}

- 5 u^{2} + \frac{2}{u ^{2}} u^{2} - 3 u^{2} u^{2} = 0 \cdot u^{2}

化简

- 5 u^{2} + \frac{2}{u ^{2}} u^{2} - 3 u^{2} u^{2} = 0 \cdot u^{2}

化简

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

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

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

约分:

u^{2}

= 2

化简

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

- 3 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}

化简

0 \cdot u^{2} : 0

0 \cdot u^{2}

使用法则

0 \cdot a = 0

= 0

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

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

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

解

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

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

改写成标准形式

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

- 3 u^{4} - 5 u^{2} + 2 = 0

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

- 3 v^{2} - 5 v + 2 = 0

解

- 3 v^{2} - 5 v + 2 = 0 : v = - 2, v = \frac{1}{3}

- 3 v^{2} - 5 v + 2 = 0

使用求根公式求解

- 3 v^{2} - 5 v + 2 = 0

二次方程求根公式:

若

a = - 3, b = - 5, c = 2

v_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 3 ) \cdot 2}{2 ( - 3 )}

v_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 3 ) \cdot 2}{2 ( - 3 )}

(- 5)^{2} - 4 (- 3) \cdot 2 = 7

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

使用法则

- (- a) = a

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

使用指数法则:

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

若

n

是偶数

(- 5)^{2} = 5^{2}

= 5^{2} + 4 \cdot 3 \cdot 2

数字相乘:

4 \cdot 3 \cdot 2 = 24

= 5^{2} + 24

5^{2} = 25

= 25 + 24

数字相加:

25 + 24 = 49

= 49

因式分解数字:

49 = 7^{2}

= 7^{2}

使用根式运算法则:

7^{2} = 7

= 7

v_{1, 2} = \frac{- ( - 5 ) \pm 7}{2 ( - 3 )}

将解分隔开

v_{1} = \frac{- ( - 5 ) + 7}{2 ( - 3 )}, v_{2} = \frac{- ( - 5 ) - 7}{2 ( - 3 )}

v = \frac{- ( - 5 ) + 7}{2 ( - 3 )} : - 2

\frac{- ( - 5 ) + 7}{2 ( - 3 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{5 + 7}{- 2 \cdot 3}

数字相加:

5 + 7 = 12

= \frac{12}{- 2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{12}{- 6}

使用分式法则:

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

= - \frac{12}{6}

数字相除:

\frac{12}{6} = 2

= - 2

v = \frac{- ( - 5 ) - 7}{2 ( - 3 )} : \frac{1}{3}

\frac{- ( - 5 ) - 7}{2 ( - 3 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{5 - 7}{- 2 \cdot 3}

数字相减:

5 - 7 = - 2

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

数字相乘:

2 \cdot 3 = 6

= \frac{- 2}{- 6}

使用分式法则:

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

= \frac{2}{6}

约分:

2

= \frac{1}{3}

二次方程组的解是:

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

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

代回

v = u^{2}

，求解

u

解

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

u^{2} = - 2

对于

x^{2} = f (a)

解为

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

u = - 2, u = - - 2

化简

- 2 : 2 i

- 2

使用根式运算法则:

- a = - 1 a

- 2 = - 1 2

= - 1 2

使用虚数运算法则:

- 1 = i

= 2 i

化简

- - 2 : - 2 i

- - 2

化简

- 2 : 2 i

- 2

使用根式运算法则:

- a = - 1 a

- 2 = - 1 2

= - 1 2

使用虚数运算法则:

- 1 = i

= 2 i

= - 2 i

u = 2 i, u = - 2 i

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

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

验证解

找到无定义的点（奇点）:

u = 0

取

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

u = cot (x)

代回

cot (x) = 2 i, cot (x) = - 2 i, cot (x) = \frac{1}{3}, cot (x) = - \frac{1}{3}

cot (x) = 2 i, cot (x) = - 2 i, cot (x) = \frac{1}{3}, cot (x) = - \frac{1}{3}

cot (x) = 2 i :

无解

cot (x) = 2 i

无解

cot (x) = - 2 i :

无解

cot (x) = - 2 i

无解

cot (x) = \frac{1}{3} : x = arccot (\frac{1}{3}) + πn

cot (x) = \frac{1}{3}

使用反三角函数性质

cot (x) = \frac{1}{3}

cot (x) = \frac{1}{3}

的通解

cot (x) = a \Rightarrow x = arccot (a) + πn

x = arccot (\frac{1}{3}) + πn

x = arccot (\frac{1}{3}) + πn

cot (x) = - \frac{1}{3} : x = arccot (- \frac{1}{3}) + πn

cot (x) = - \frac{1}{3}

使用反三角函数性质

cot (x) = - \frac{1}{3}

cot (x) = - \frac{1}{3}

的通解

cot (x) = - a \Rightarrow x = arccot (- a) + πn

x = arccot (- \frac{1}{3}) + πn

x = arccot (- \frac{1}{3}) + πn

合并所有解

x = arccot (\frac{1}{3}) + πn, x = arccot (- \frac{1}{3}) + πn

以小数形式表示解

x = 1.04719 \dots + πn, x = 2.09439 \dots + πn

作图

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