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受欢迎的三角函数 >

cot(x)+tan(x)=1

解答

cot (x) + tan (x) = 1

解答

x \in R 无解

求解步骤

cot (x) + tan (x) = 1

两边减去

1

cot (x) + tan (x) - 1 = 0

使用三角恒等式改写

- 1 + cot (x) + tan (x)

使用基本三角恒等式:

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

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

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

用替代法求解

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

令

: cot (x) = u

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

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

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

在两边乘以

u

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

在两边乘以

u

- 1 \cdot u + uu + \frac{1}{u} u = 0 \cdot u

化简

- 1 \cdot u + uu + \frac{1}{u} u = 0 \cdot u

化简

- 1 \cdot u : - u

- 1 \cdot u

乘以:

1 \cdot u = u

= - u

化简

uu : u^{2}

uu

使用指数法则:

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

uu = u^{1 + 1}

= u^{1 + 1}

数字相加:

1 + 1 = 2

= u^{2}

化简

\frac{1}{u} u : 1

\frac{1}{u} u

分式相乘:

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

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

约分:

u

= 1

化简

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

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

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

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

解

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

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

改写成标准形式

a x^{2} + b x + c = 0

u^{2} - u + 1 = 0

使用求根公式求解

u^{2} - u + 1 = 0

二次方程求根公式:

若

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

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

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

化简

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

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

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

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

若

n

是偶数

(- 1)^{2} = 1^{2}

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 4

= 1 - 4

数字相减:

1 - 4 = - 3

= - 3

使用根式运算法则:

- a = - 1 a

- 3 = - 1 3

= - 1 3

使用虚数运算法则:

- 1 = i

= 3 i

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

将解分隔开

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

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

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

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 1 = 2

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

将

\frac{1 + 3 i}{2}

改写成标准复数形式:

\frac{1}{2} + \frac{3}{2} i

\frac{1 + 3 i}{2}

使用分式法则:

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

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

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

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

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

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

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 1 = 2

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

将

\frac{1 - 3 i}{2}

改写成标准复数形式:

\frac{1}{2} - \frac{3}{2} i

\frac{1 - 3 i}{2}

使用分式法则:

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

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

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

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

二次方程组的解是:

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

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

u = cot (x)

代回

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

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

cot (x) = \frac{1}{2} + i \frac{3}{2} :

无解

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

无解

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

无解

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

无解

合并所有解

x \in R 无解

作图

流行的例子

2cos(x)-3sin(x)=-2

tan(x)=2-cot(x)

sin(θ)=cos(37)

cot(x)-cot(x)csc(x)=0