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受欢迎的三角函数 >

sec^2(x)+8tan(x)=10

解答

sec^{2} (x) + 8 tan (x) = 10

解答

x = \frac{π}{4} + πn, x = - 1.46013 \dots + πn

+1

度数

x = 4 5^{\circ} + 18 0^{\circ} n, x = - 83.65980 \dots^{\circ} + 18 0^{\circ} n

求解步骤

sec^{2} (x) + 8 tan (x) = 10

两边减去

10

sec^{2} (x) + 8 tan (x) - 10 = 0

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

= - 10 + tan^{2} (x) + 1 + 8 tan (x)

化简

- 10 + tan^{2} (x) + 1 + 8 tan (x) : tan^{2} (x) + 8 tan (x) - 9

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

对同类项分组

= tan^{2} (x) + 8 tan (x) - 10 + 1

数字相加/相减:

- 10 + 1 = - 9

= tan^{2} (x) + 8 tan (x) - 9

= tan^{2} (x) + 8 tan (x) - 9

- 9 + tan^{2} (x) + 8 tan (x) = 0

用替代法求解

- 9 + tan^{2} (x) + 8 tan (x) = 0

令

: tan (x) = u

- 9 + u^{2} + 8 u = 0

- 9 + u^{2} + 8 u = 0 : u = 1, u = - 9

- 9 + u^{2} + 8 u = 0

改写成标准形式

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

u^{2} + 8 u - 9 = 0

使用求根公式求解

u^{2} + 8 u - 9 = 0

二次方程求根公式:

若

a = 1, b = 8, c = - 9

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot ( - 9 )}{2 \cdot 1}

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot ( - 9 )}{2 \cdot 1}

8^{2} - 4 \cdot 1 \cdot (- 9) = 10

8^{2} - 4 \cdot 1 \cdot (- 9)

使用法则

- (- a) = a

= 8^{2} + 4 \cdot 1 \cdot 9

数字相乘:

4 \cdot 1 \cdot 9 = 36

= 8^{2} + 36

8^{2} = 64

= 64 + 36

数字相加:

64 + 36 = 100

= 100

因式分解数字:

100 = 1 0^{2}

= 1 0^{2}

使用根式运算法则:

n a^{n} = a

1 0^{2} = 10

= 10

u_{1, 2} = \frac{- 8 \pm 10}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- 8 + 10}{2 \cdot 1}, u_{2} = \frac{- 8 - 10}{2 \cdot 1}

u = \frac{- 8 + 10}{2 \cdot 1} : 1

\frac{- 8 + 10}{2 \cdot 1}

数字相加/相减:

- 8 + 10 = 2

= \frac{2}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{2}{2}

使用法则

\frac{a}{a} = 1

= 1

u = \frac{- 8 - 10}{2 \cdot 1} : - 9

\frac{- 8 - 10}{2 \cdot 1}

数字相减:

- 8 - 10 = - 18

= \frac{- 18}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 18}{2}

使用分式法则:

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

= - \frac{18}{2}

数字相除:

\frac{18}{2} = 9

= - 9

二次方程组的解是:

u = 1, u = - 9

u = tan (x)

代回

tan (x) = 1, tan (x) = - 9

tan (x) = 1, tan (x) = - 9

tan (x) = 1 : x = \frac{π}{4} + πn

tan (x) = 1

tan (x) = 1

的通解

tan (x)

周期表（周期为

πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

tan (x) = - 9 : x = arctan (- 9) + πn

tan (x) = - 9

使用反三角函数性质

tan (x) = - 9

tan (x) = - 9

的通解

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

x = arctan (- 9) + πn

x = arctan (- 9) + πn

合并所有解

x = \frac{π}{4} + πn, x = arctan (- 9) + πn

以小数形式表示解

x = \frac{π}{4} + πn, x = - 1.46013 \dots + πn

作图

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