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受欢迎的三角函数 >

4tan^2(x)+7tan(x)-4=0,cot(2x)

解答

4 tan^{2} (x) + 7 tan (x) - 4 = 0, cot (2 x)

解答

x \in R 无解

求解步骤

4 tan^{2} (x) + 7 tan (x) - 4 = 0, cot (2 x)

用替代法求解

4 tan^{2} (x) + 7 tan (x) - 4 = 0

令

: tan (x) = u

4 u^{2} + 7 u - 4 = 0

4 u^{2} + 7 u - 4 = 0 : u = \frac{- 7 + 113}{8}, u = \frac{- 7 - 113}{8}

4 u^{2} + 7 u - 4 = 0

使用求根公式求解

4 u^{2} + 7 u - 4 = 0

二次方程求根公式:

若

a = 4, b = 7, c = - 4

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 4 ( - 4 )}{2 \cdot 4}

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 4 ( - 4 )}{2 \cdot 4}

7^{2} - 4 \cdot 4 (- 4) = 113

7^{2} - 4 \cdot 4 (- 4)

使用法则

- (- a) = a

= 7^{2} + 4 \cdot 4 \cdot 4

数字相乘:

4 \cdot 4 \cdot 4 = 64

= 7^{2} + 64

7^{2} = 49

= 49 + 64

数字相加:

49 + 64 = 113

= 113

u_{1, 2} = \frac{- 7 \pm 113}{2 \cdot 4}

将解分隔开

u_{1} = \frac{- 7 + 113}{2 \cdot 4}, u_{2} = \frac{- 7 - 113}{2 \cdot 4}

u = \frac{- 7 + 113}{2 \cdot 4} : \frac{- 7 + 113}{8}

\frac{- 7 + 113}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 7 + 113}{8}

u = \frac{- 7 - 113}{2 \cdot 4} : \frac{- 7 - 113}{8}

\frac{- 7 - 113}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 7 - 113}{8}

二次方程组的解是:

u = \frac{- 7 + 113}{8}, u = \frac{- 7 - 113}{8}

u = tan (x)

代回

tan (x) = \frac{- 7 + 113}{8}, tan (x) = \frac{- 7 - 113}{8}

tan (x) = \frac{- 7 + 113}{8}, tan (x) = \frac{- 7 - 113}{8}

tan (x) = \frac{- 7 + 113}{8}, cot (2 x) :

无解

tan (x) = \frac{- 7 + 113}{8}, cot (2 x)

使用反三角函数性质

tan (x) = \frac{- 7 + 113}{8}

tan (x) = \frac{- 7 + 113}{8}

的通解

tan (x) = a \Rightarrow x = arctan (a) + πn

x = arctan (\frac{- 7 + 113}{8}) + πn

x = arctan (\frac{- 7 + 113}{8}) + πn

在

cot (2 x)

范围内的解

无解

tan (x) = \frac{- 7 - 113}{8}, cot (2 x) :

无解

tan (x) = \frac{- 7 - 113}{8}, cot (2 x)

使用反三角函数性质

tan (x) = \frac{- 7 - 113}{8}

tan (x) = \frac{- 7 - 113}{8}

的通解

tan (x) = - a \Rightarrow x = arctan (- a) + πn

x = arctan (\frac{- 7 - 113}{8}) + πn

x = arctan (\frac{- 7 - 113}{8}) + πn

在

cot (2 x)

范围内的解

无解

合并所有解

x \in R 无解

作图

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