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受欢迎的三角函数 >

2sec^2(x)+4tan(x)=1

解答

2 sec^{2} (x) + 4 tan (x) = 1

解答

x = - 0.28492 \dots + πn, x = - 1.04089 \dots + πn

+1

度数

x = - 16.32494 \dots^{\circ} + 18 0^{\circ} n, x = - 59.63880 \dots^{\circ} + 18 0^{\circ} n

求解步骤

2 sec^{2} (x) + 4 tan (x) = 1

两边减去

1

2 sec^{2} (x) + 4 tan (x) - 1 = 0

使用三角恒等式改写

- 1 + 2 sec^{2} (x) + 4 tan (x)

使用毕达哥拉斯恒等式:

sec^{2} (x) = tan^{2} (x) + 1

= - 1 + 2 (tan^{2} (x) + 1) + 4 tan (x)

化简

- 1 + 2 (tan^{2} (x) + 1) + 4 tan (x) : 2 tan^{2} (x) + 4 tan (x) + 1

- 1 + 2 (tan^{2} (x) + 1) + 4 tan (x)

乘开

2 (tan^{2} (x) + 1) : 2 tan^{2} (x) + 2

2 (tan^{2} (x) + 1)

使用分配律:

a (b + c) = ab + a c

a = 2, b = tan^{2} (x), c = 1

= 2 tan^{2} (x) + 2 \cdot 1

数字相乘:

2 \cdot 1 = 2

= 2 tan^{2} (x) + 2

= - 1 + 2 tan^{2} (x) + 2 + 4 tan (x)

化简

- 1 + 2 tan^{2} (x) + 2 + 4 tan (x) : 2 tan^{2} (x) + 4 tan (x) + 1

- 1 + 2 tan^{2} (x) + 2 + 4 tan (x)

对同类项分组

= 2 tan^{2} (x) + 4 tan (x) - 1 + 2

数字相加/相减:

- 1 + 2 = 1

= 2 tan^{2} (x) + 4 tan (x) + 1

= 2 tan^{2} (x) + 4 tan (x) + 1

= 2 tan^{2} (x) + 4 tan (x) + 1

1 + 2 tan^{2} (x) + 4 tan (x) = 0

用替代法求解

1 + 2 tan^{2} (x) + 4 tan (x) = 0

令

: tan (x) = u

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

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = 2, b = 4, c = 1

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

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

4^{2} - 4 \cdot 2 \cdot 1 = 22

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

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 4^{2} - 8

4^{2} = 16

= 16 - 8

数字相减:

16 - 8 = 8

= 8

8

质因数分解:

2^{3}

8

8

除以

28 = 4 \cdot 2

= 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

使用指数法则:

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

= 2^{2} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 22

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

将解分隔开

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

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

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

数字相乘:

2 \cdot 2 = 4

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

分解

- 4 + 22 : 2 (- 2 + 2)

- 4 + 22

改写为

= - 2 \cdot 2 + 22

因式分解出通项

2

= 2 (- 2 + 2)

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

约分:

2

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

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

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

数字相乘:

2 \cdot 2 = 4

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

分解

- 4 - 22 : - 2 (2 + 2)

- 4 - 22

改写为

= - 2 \cdot 2 - 22

因式分解出通项

2

= - 2 (2 + 2)

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

约分:

2

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

二次方程组的解是:

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

u = tan (x)

代回

tan (x) = \frac{- 2 + 2}{2}, tan (x) = - \frac{2 + 2}{2}

tan (x) = \frac{- 2 + 2}{2}, tan (x) = - \frac{2 + 2}{2}

tan (x) = \frac{- 2 + 2}{2} : x = arctan (\frac{- 2 + 2}{2}) + πn

tan (x) = \frac{- 2 + 2}{2}

使用反三角函数性质

tan (x) = \frac{- 2 + 2}{2}

tan (x) = \frac{- 2 + 2}{2}

的通解

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

x = arctan (\frac{- 2 + 2}{2}) + πn

x = arctan (\frac{- 2 + 2}{2}) + πn

tan (x) = - \frac{2 + 2}{2} : x = arctan (- \frac{2 + 2}{2}) + πn

tan (x) = - \frac{2 + 2}{2}

使用反三角函数性质

tan (x) = - \frac{2 + 2}{2}

tan (x) = - \frac{2 + 2}{2}

的通解

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

x = arctan (- \frac{2 + 2}{2}) + πn

x = arctan (- \frac{2 + 2}{2}) + πn

合并所有解

x = arctan (\frac{- 2 + 2}{2}) + πn, x = arctan (- \frac{2 + 2}{2}) + πn

以小数形式表示解

x = - 0.28492 \dots + πn, x = - 1.04089 \dots + πn

作图

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