zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

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

解答

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

解答

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

+1

度数

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

求解步骤

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

使用三角恒等式改写

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

使用基本三角恒等式:

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

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

- 2 + cot^{2} (x) + \frac{1}{cot ^{2} ( x )} = 0

用替代法求解

- 2 + cot^{2} (x) + \frac{1}{cot ^{2} ( x )} = 0

令

: cot (x) = u

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

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

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

在两边乘以

u^{2}

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

在两边乘以

u^{2}

- 2 u^{2} + u^{2} u^{2} + \frac{1}{u ^{2}} u^{2} = 0 \cdot u^{2}

化简

- 2 u^{2} + u^{2} u^{2} + \frac{1}{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{1}{u ^{2}} u^{2} : 1

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

分式相乘:

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

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

约分:

u^{2}

= 1

化简

0 \cdot u^{2} : 0

0 \cdot u^{2}

使用法则

0 \cdot a = 0

= 0

- 2 u^{2} + u^{4} + 1 = 0

- 2 u^{2} + u^{4} + 1 = 0

- 2 u^{2} + u^{4} + 1 = 0

解

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

- 2 u^{2} + u^{4} + 1 = 0

改写成标准形式

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

u^{4} - 2 u^{2} + 1 = 0

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

v^{2} - 2 v + 1 = 0

解

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

v^{2} - 2 v + 1 = 0

使用求根公式求解

v^{2} - 2 v + 1 = 0

二次方程求根公式:

若

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

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

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

(- 2)^{2} - 4 \cdot 1 \cdot 1 = 0

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

使用指数法则:

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

若

n

是偶数

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

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

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 2^{2} - 4

2^{2} = 4

= 4 - 4

数字相减:

4 - 4 = 0

= 0

v_{1, 2} = \frac{- ( - 2 ) \pm 0}{2 \cdot 1}

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

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

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

使用法则

- (- a) = a

= \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

v = 1

二次方程组的解是:

v = 1

v = 1

代回

v = u^{2}

，求解

u

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 = 1, u = - 1

u = 1, u = - 1

验证解

找到无定义的点（奇点）:

u = 0

取

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

u = cot (x)

代回

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

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

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

作图

流行的例子

(1+2cos(2x))^2=0

(1 + 2 cos (2 x))^{2} = 0

tan (2 x) = 0.6

2 sin (3 θ) = - 1

cos((pi(x-7))/3)= 1/2

cos (\frac{π ( x - 7 )}{3}) = \frac{1}{2}

2cos(2x)+8cos(x)+5=0

2 cos (2 x) + 8 cos (x) + 5 = 0