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受欢迎的三角函数 >

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

解答

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

解答

x \in R 无解

求解步骤

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

两边减去

- 2

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

使用三角恒等式改写

2 + tan (x) + 2 cot (x)

使用基本三角恒等式:

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

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

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

用替代法求解

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

令

: cot (x) = u

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

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

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

在两边乘以

u

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

在两边乘以

u

2 u + \frac{1}{u} u + 2 uu = 0 \cdot u

化简

2 u + \frac{1}{u} u + 2 uu = 0 \cdot u

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

2 uu : 2 u^{2}

2 uu

使用指数法则:

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

uu = u^{1 + 1}

= 2 u^{1 + 1}

数字相加:

1 + 1 = 2

= 2 u^{2}

化简

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

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

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

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

解

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = 2, b = 2, c = 1

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

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

化简

2^{2} - 4 \cdot 2 \cdot 1 : 2 i

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

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 2^{2} - 8

使用虚数运算法则:

- a = i a

= i 8 - 2^{2}

- 2^{2} + 8 = 2

- 2^{2} + 8

2^{2} = 4

= - 4 + 8

数字相加/相减:

- 4 + 8 = 4

= 4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= 2 i

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

将解分隔开

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

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

\frac{- 2 + 2 i}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 2 + 2 i}{4}

分解

- 2 + 2 i : 2 (- 1 + i)

- 2 + 2 i

改写为

= - 2 \cdot 1 + 2 i

因式分解出通项

2

= 2 (- 1 + i)

= \frac{2 ( - 1 + i )}{4}

约分:

2

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

将

\frac{- 1 + i}{2}

改写成标准复数形式:

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

\frac{- 1 + i}{2}

使用分式法则:

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

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

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

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

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

\frac{- 2 - 2 i}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 2 - 2 i}{4}

分解

- 2 - 2 i : - 2 (1 + i)

- 2 - 2 i

改写为

= - 2 \cdot 1 - 2 i

因式分解出通项

2

= - 2 (1 + i)

= - \frac{2 ( 1 + i )}{4}

约分:

2

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

将

- \frac{1 + i}{2}

改写成标准复数形式:

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

- \frac{1 + i}{2}

使用分式法则:

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

\frac{1 + i}{2} = - (\frac{1}{2}) - (\frac{i}{2})

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

去除括号:

(a) = a

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

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

二次方程组的解是:

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

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

u = cot (x)

代回

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

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

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

无解

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

无解

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

无解

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

无解

合并所有解

x \in R 无解

作图

流行的例子

sin(x)-sqrt(3)=0

sin (x) - 3 = 0

2 tan (4 x) = 1

cos(θ)=(5713634.83)/((2236)(3733))

cos (θ) = \frac{5713634.83}{( 2236 ) ( 3733 )}

sin(x)= 3/(sqrt(58))

sin (x) = \frac{3}{58}

cos (2 x + 1) = 0