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受欢迎的三角函数 >

2sec(x)+tan^2(x)-9=1

解答

2 sec (x) + tan^{2} (x) - 9 = 1

解答

x = 1.15291 \dots + 2 πn, x = 2 π - 1.15291 \dots + 2 πn, x = 1.79672 \dots + 2 πn, x = - 1.79672 \dots + 2 πn

+1

度数

x = 66.05701 \dots^{\circ} + 36 0^{\circ} n, x = 293.94298 \dots^{\circ} + 36 0^{\circ} n, x = 102.94462 \dots^{\circ} + 36 0^{\circ} n, x = - 102.94462 \dots^{\circ} + 36 0^{\circ} n

求解步骤

2 sec (x) + tan^{2} (x) - 9 = 1

两边减去

1

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

化简

- 10 + sec^{2} (x) - 1 + 2 sec (x) : sec^{2} (x) + 2 sec (x) - 11

- 10 + sec^{2} (x) - 1 + 2 sec (x)

对同类项分组

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

数字相减:

- 10 - 1 = - 11

= sec^{2} (x) + 2 sec (x) - 11

= sec^{2} (x) + 2 sec (x) - 11

- 11 + sec^{2} (x) + 2 sec (x) = 0

用替代法求解

- 11 + sec^{2} (x) + 2 sec (x) = 0

令

: sec (x) = u

- 11 + u^{2} + 2 u = 0

- 11 + u^{2} + 2 u = 0 : u = - 1 + 23, u = - 1 - 23

- 11 + u^{2} + 2 u = 0

改写成标准形式

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

u^{2} + 2 u - 11 = 0

使用求根公式求解

u^{2} + 2 u - 11 = 0

二次方程求根公式:

若

a = 1, b = 2, c = - 11

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 11 )}{2 \cdot 1}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 11 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 11) = 43

2^{2} - 4 \cdot 1 \cdot (- 11)

使用法则

- (- a) = a

= 2^{2} + 4 \cdot 1 \cdot 11

数字相乘:

4 \cdot 1 \cdot 11 = 44

= 2^{2} + 44

2^{2} = 4

= 4 + 44

数字相加:

4 + 44 = 48

= 48

48

质因数分解:

2^{4} \cdot 3

48

48

除以

248 = 24 \cdot 2

= 2 \cdot 24

24

除以

224 = 12 \cdot 2

= 2 \cdot 2 \cdot 12

12

除以

212 = 6 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 6

6

除以

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3

= 2^{4} \cdot 3

= 2^{4} \cdot 3

使用根式运算法则:

n ab = n a n b

= 3 2^{4}

使用根式运算法则:

n a^{m} = a^{\frac{m}{n}}

2^{4} = 2^{\frac{4}{2}} = 2^{2}

= 2^{2} 3

整理后得

= 43

u_{1, 2} = \frac{- 2 \pm 4 3}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- 2 + 4 3}{2 \cdot 1}, u_{2} = \frac{- 2 - 4 3}{2 \cdot 1}

u = \frac{- 2 + 4 3}{2 \cdot 1} : - 1 + 23

\frac{- 2 + 4 3}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 2 + 4 3}{2}

分解

- 2 + 43 : 2 (- 1 + 23)

- 2 + 43

改写为

= - 2 \cdot 1 + 2 \cdot 23

因式分解出通项

2

= 2 (- 1 + 23)

= \frac{2 ( - 1 + 2 3 )}{2}

数字相除:

\frac{2}{2} = 1

= - 1 + 23

u = \frac{- 2 - 4 3}{2 \cdot 1} : - 1 - 23

\frac{- 2 - 4 3}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 2 - 4 3}{2}

分解

- 2 - 43 : - 2 (1 + 23)

- 2 - 43

改写为

= - 2 \cdot 1 - 2 \cdot 23

因式分解出通项

2

= - 2 (1 + 23)

= - \frac{2 ( 1 + 2 3 )}{2}

数字相除:

\frac{2}{2} = 1

= - (1 + 23)

取负

- (1 + 23) = - 1 - 23

= - 1 - 23

二次方程组的解是:

u = - 1 + 23, u = - 1 - 23

u = sec (x)

代回

sec (x) = - 1 + 23, sec (x) = - 1 - 23

sec (x) = - 1 + 23, sec (x) = - 1 - 23

sec (x) = - 1 + 23 : x = arcsec (- 1 + 23) + 2 πn, x = 2 π - arcsec (- 1 + 23) + 2 πn

sec (x) = - 1 + 23

使用反三角函数性质

sec (x) = - 1 + 23

sec (x) = - 1 + 23

的通解

sec (x) = a \Rightarrow x = arcsec (a) + 2 πn, x = 2 π - arcsec (a) + 2 πn

x = arcsec (- 1 + 23) + 2 πn, x = 2 π - arcsec (- 1 + 23) + 2 πn

x = arcsec (- 1 + 23) + 2 πn, x = 2 π - arcsec (- 1 + 23) + 2 πn

sec (x) = - 1 - 23 : x = arcsec (- 1 - 23) + 2 πn, x = - arcsec (- 1 - 23) + 2 πn

sec (x) = - 1 - 23

使用反三角函数性质

sec (x) = - 1 - 23

sec (x) = - 1 - 23

的通解

sec (x) = - a \Rightarrow x = arcsec (- a) + 2 πn, x = - arcsec (- a) + 2 πn

x = arcsec (- 1 - 23) + 2 πn, x = - arcsec (- 1 - 23) + 2 πn

x = arcsec (- 1 - 23) + 2 πn, x = - arcsec (- 1 - 23) + 2 πn

合并所有解

x = arcsec (- 1 + 23) + 2 πn, x = 2 π - arcsec (- 1 + 23) + 2 πn, x = arcsec (- 1 - 23) + 2 πn, x = - arcsec (- 1 - 23) + 2 πn

以小数形式表示解

x = 1.15291 \dots + 2 πn, x = 2 π - 1.15291 \dots + 2 πn, x = 1.79672 \dots + 2 πn, x = - 1.79672 \dots + 2 πn

作图

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