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受欢迎的三角函数 >

5sec^2(x)-14tan(x)=8

解答

5 sec^{2} (x) - 14 tan (x) = 8

解答

x = 1.24904 \dots + πn, x = - 0.19739 \dots + πn

+1

度数

x = 71.56505 \dots^{\circ} + 18 0^{\circ} n, x = - 11.30993 \dots^{\circ} + 18 0^{\circ} n

求解步骤

5 sec^{2} (x) - 14 tan (x) = 8

两边减去

8

5 sec^{2} (x) - 14 tan (x) - 8 = 0

使用三角恒等式改写

- 8 - 14 tan (x) + 5 sec^{2} (x)

使用毕达哥拉斯恒等式:

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

= - 8 - 14 tan (x) + 5 (tan^{2} (x) + 1)

化简

- 8 - 14 tan (x) + 5 (tan^{2} (x) + 1) : 5 tan^{2} (x) - 14 tan (x) - 3

- 8 - 14 tan (x) + 5 (tan^{2} (x) + 1)

乘开

5 (tan^{2} (x) + 1) : 5 tan^{2} (x) + 5

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

使用分配律:

a (b + c) = ab + a c

a = 5, b = tan^{2} (x), c = 1

= 5 tan^{2} (x) + 5 \cdot 1

数字相乘:

5 \cdot 1 = 5

= 5 tan^{2} (x) + 5

= - 8 - 14 tan (x) + 5 tan^{2} (x) + 5

化简

- 8 - 14 tan (x) + 5 tan^{2} (x) + 5 : 5 tan^{2} (x) - 14 tan (x) - 3

- 8 - 14 tan (x) + 5 tan^{2} (x) + 5

对同类项分组

= - 14 tan (x) + 5 tan^{2} (x) - 8 + 5

数字相加/相减:

- 8 + 5 = - 3

= 5 tan^{2} (x) - 14 tan (x) - 3

= 5 tan^{2} (x) - 14 tan (x) - 3

= 5 tan^{2} (x) - 14 tan (x) - 3

- 3 - 14 tan (x) + 5 tan^{2} (x) = 0

用替代法求解

- 3 - 14 tan (x) + 5 tan^{2} (x) = 0

令

: tan (x) = u

- 3 - 14 u + 5 u^{2} = 0

- 3 - 14 u + 5 u^{2} = 0 : u = 3, u = - \frac{1}{5}

- 3 - 14 u + 5 u^{2} = 0

改写成标准形式

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

5 u^{2} - 14 u - 3 = 0

使用求根公式求解

5 u^{2} - 14 u - 3 = 0

二次方程求根公式:

若

a = 5, b = - 14, c = - 3

u_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 5 ( - 3 )}{2 \cdot 5}

u_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 5 ( - 3 )}{2 \cdot 5}

(- 14)^{2} - 4 \cdot 5 (- 3) = 16

(- 14)^{2} - 4 \cdot 5 (- 3)

使用法则

- (- a) = a

= (- 14)^{2} + 4 \cdot 5 \cdot 3

使用指数法则:

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

若

n

是偶数

(- 14)^{2} = 1 4^{2}

= 1 4^{2} + 4 \cdot 5 \cdot 3

数字相乘:

4 \cdot 5 \cdot 3 = 60

= 1 4^{2} + 60

1 4^{2} = 196

= 196 + 60

数字相加:

196 + 60 = 256

= 256

因式分解数字:

256 = 1 6^{2}

= 1 6^{2}

使用根式运算法则:

n a^{n} = a

1 6^{2} = 16

= 16

u_{1, 2} = \frac{- ( - 14 ) \pm 16}{2 \cdot 5}

将解分隔开

u_{1} = \frac{- ( - 14 ) + 16}{2 \cdot 5}, u_{2} = \frac{- ( - 14 ) - 16}{2 \cdot 5}

u = \frac{- ( - 14 ) + 16}{2 \cdot 5} : 3

\frac{- ( - 14 ) + 16}{2 \cdot 5}

使用法则

- (- a) = a

= \frac{14 + 16}{2 \cdot 5}

数字相加:

14 + 16 = 30

= \frac{30}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{30}{10}

数字相除:

\frac{30}{10} = 3

= 3

u = \frac{- ( - 14 ) - 16}{2 \cdot 5} : - \frac{1}{5}

\frac{- ( - 14 ) - 16}{2 \cdot 5}

使用法则

- (- a) = a

= \frac{14 - 16}{2 \cdot 5}

数字相减:

14 - 16 = - 2

= \frac{- 2}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{- 2}{10}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2}{10}

约分:

2

= - \frac{1}{5}

二次方程组的解是:

u = 3, u = - \frac{1}{5}

u = tan (x)

代回

tan (x) = 3, tan (x) = - \frac{1}{5}

tan (x) = 3, tan (x) = - \frac{1}{5}

tan (x) = 3 : x = arctan (3) + πn

tan (x) = 3

使用反三角函数性质

tan (x) = 3

tan (x) = 3

的通解

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

x = arctan (3) + πn

x = arctan (3) + πn

tan (x) = - \frac{1}{5} : x = arctan (- \frac{1}{5}) + πn

tan (x) = - \frac{1}{5}

使用反三角函数性质

tan (x) = - \frac{1}{5}

tan (x) = - \frac{1}{5}

的通解

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

x = arctan (- \frac{1}{5}) + πn

x = arctan (- \frac{1}{5}) + πn

合并所有解

x = arctan (3) + πn, x = arctan (- \frac{1}{5}) + πn

以小数形式表示解

x = 1.24904 \dots + πn, x = - 0.19739 \dots + πn

作图

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