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受欢迎的三角函数 >

2tan^2(x)+cot^2(x)-3=0

解答

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

解答

x = 0.61547 \dots + πn, x = 2.52611 \dots + πn, x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

+1

度数

x = 35.26438 \dots^{\circ} + 18 0^{\circ} n, x = 144.73561 \dots^{\circ} + 18 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n

求解步骤

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

使用三角恒等式改写

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

使用基本三角恒等式:

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

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

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 )}

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

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

用替代法求解

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

令

: cot (x) = u

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

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

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

在两边乘以

u^{2}

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

在两边乘以

u^{2}

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

化简

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

化简

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

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}

= u^{2 + 2}

数字相加:

2 + 2 = 4

= u^{4}

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

0 \cdot u^{2} : 0

0 \cdot u^{2}

使用法则

0 \cdot a = 0

= 0

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

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

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

解

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

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

改写成标准形式

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

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

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

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

解

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

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

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

使用指数法则:

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

若

n

是偶数

(- 3)^{2} = 3^{2}

= 3^{2} - 4 \cdot 1 \cdot 2

数字相乘:

4 \cdot 1 \cdot 2 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

数字相减:

9 - 8 = 1

= 1

使用法则

1 = 1

= 1

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

将解分隔开

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

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

\frac{- ( - 3 ) + 1}{2 \cdot 1}

使用法则

- (- a) = a

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

数字相加:

3 + 1 = 4

= \frac{4}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{4}{2}

数字相除:

\frac{4}{2} = 2

= 2

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

\frac{- ( - 3 ) - 1}{2 \cdot 1}

使用法则

- (- a) = a

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

数字相减:

3 - 1 = 2

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

数字相乘:

2 \cdot 1 = 2

= \frac{2}{2}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

v = 2, v = 1

v = 2, v = 1

代回

v = u^{2}

，求解

u

解

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

u^{2} = 2

对于

x^{2} = f (a)

解为

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

u = 2, u = - 2

解

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

u^{2} = 1

对于

x^{2} = f (a)

解为

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

u = 1, u = - 1

1 = 1

1

使用法则

1 = 1

= 1

- 1 = - 1

- 1

使用法则

1 = 1

= - 1

u = 1, u = - 1

解为

u = 2, u = - 2, u = 1, u = - 1

u = 2, u = - 2, u = 1, u = - 1

验证解

找到无定义的点（奇点）:

u = 0

取

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

u = cot (x)

代回

cot (x) = 2, cot (x) = - 2, cot (x) = 1, cot (x) = - 1

cot (x) = 2, cot (x) = - 2, cot (x) = 1, cot (x) = - 1

cot (x) = 2 : x = arccot (2) + πn

cot (x) = 2

使用反三角函数性质

cot (x) = 2

cot (x) = 2

的通解

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

x = arccot (2) + πn

x = arccot (2) + πn

cot (x) = - 2 : x = arccot (- 2) + πn

cot (x) = - 2

使用反三角函数性质

cot (x) = - 2

cot (x) = - 2

的通解

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

x = arccot (- 2) + πn

x = arccot (- 2) + πn

cot (x) = 1 : x = \frac{π}{4} + πn

cot (x) = 1

cot (x) = 1

的通解

cot (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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

cot (x) = - 1 : x = \frac{3 π}{4} + πn

cot (x) = - 1

cot (x) = - 1

的通解

cot (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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

合并所有解

x = arccot (2) + πn, x = arccot (- 2) + πn, x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

以小数形式表示解

x = 0.61547 \dots + πn, x = 2.52611 \dots + πn, x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

作图

流行的例子

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