zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

tan^2(x)+7=-10sec(x),0<x<2pi

解答

tan^{2} (x) + 7 = - 10 sec (x), 0 < x < 2 π

解答

x = 1.67785 \dots, x = - 1.67785 \dots + 2 π

+1

度数

x = 96.13377 \dots^{\circ}, x = 263.86622 \dots^{\circ}

求解步骤

tan^{2} (x) + 7 = - 10 sec (x), 0 < x < 2 π

两边减去

- 10 sec (x)

tan^{2} (x) + 7 + 10 sec (x) = 0

使用三角恒等式改写

7 + tan^{2} (x) + 10 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

= 7 + sec^{2} (x) - 1 + 10 sec (x)

化简

7 + sec^{2} (x) - 1 + 10 sec (x) : sec^{2} (x) + 10 sec (x) + 6

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

对同类项分组

= sec^{2} (x) + 10 sec (x) + 7 - 1

数字相加/相减:

7 - 1 = 6

= sec^{2} (x) + 10 sec (x) + 6

= sec^{2} (x) + 10 sec (x) + 6

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

用替代法求解

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

令

: sec (x) = u

6 + u^{2} + 10 u = 0

6 + u^{2} + 10 u = 0 : u = - 5 + 19, u = - 5 - 19

6 + u^{2} + 10 u = 0

改写成标准形式

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

u^{2} + 10 u + 6 = 0

使用求根公式求解

u^{2} + 10 u + 6 = 0

二次方程求根公式:

若

a = 1, b = 10, c = 6

u_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 1 \cdot 6}{2 \cdot 1}

u_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 1 \cdot 6}{2 \cdot 1}

1 0^{2} - 4 \cdot 1 \cdot 6 = 219

1 0^{2} - 4 \cdot 1 \cdot 6

数字相乘:

4 \cdot 1 \cdot 6 = 24

= 1 0^{2} - 24

1 0^{2} = 100

= 100 - 24

数字相减:

100 - 24 = 76

= 76

76

质因数分解:

2^{2} \cdot 19

76

76

除以

276 = 38 \cdot 2

= 2 \cdot 38

38

除以

238 = 19 \cdot 2

= 2 \cdot 2 \cdot 19

2, 19

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 19

= 2^{2} \cdot 19

= 2^{2} \cdot 19

使用根式运算法则:

n ab = n a n b

= 19 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 219

u_{1, 2} = \frac{- 10 \pm 2 19}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- 10 + 2 19}{2 \cdot 1}, u_{2} = \frac{- 10 - 2 19}{2 \cdot 1}

u = \frac{- 10 + 2 19}{2 \cdot 1} : - 5 + 19

\frac{- 10 + 2 19}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 10 + 2 19}{2}

分解

- 10 + 219 : 2 (- 5 + 19)

- 10 + 219

改写为

= - 2 \cdot 5 + 219

因式分解出通项

2

= 2 (- 5 + 19)

= \frac{2 ( - 5 + 19 )}{2}

数字相除:

\frac{2}{2} = 1

= - 5 + 19

u = \frac{- 10 - 2 19}{2 \cdot 1} : - 5 - 19

\frac{- 10 - 2 19}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 10 - 2 19}{2}

分解

- 10 - 219 : - 2 (5 + 19)

- 10 - 219

改写为

= - 2 \cdot 5 - 219

因式分解出通项

2

= - 2 (5 + 19)

= - \frac{2 ( 5 + 19 )}{2}

数字相除:

\frac{2}{2} = 1

= - (5 + 19)

取负

- (5 + 19) = - 5 - 19

= - 5 - 19

二次方程组的解是:

u = - 5 + 19, u = - 5 - 19

u = sec (x)

代回

sec (x) = - 5 + 19, sec (x) = - 5 - 19

sec (x) = - 5 + 19, sec (x) = - 5 - 19

sec (x) = - 5 + 19, 0 < x < 2 π :

无解

sec (x) = - 5 + 19, 0 < x < 2 π

sec (x) \leq - 1 or sec (x) \geq 1

无解

sec (x) = - 5 - 19, 0 < x < 2 π : x = arcsec (- 5 - 19), x = - arcsec (- 5 - 19) + 2 π

sec (x) = - 5 - 19, 0 < x < 2 π

使用反三角函数性质

sec (x) = - 5 - 19

sec (x) = - 5 - 19

的通解

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

x = arcsec (- 5 - 19) + 2 πn, x = - arcsec (- 5 - 19) + 2 πn

x = arcsec (- 5 - 19) + 2 πn, x = - arcsec (- 5 - 19) + 2 πn

在

0 < x < 2 π

范围内的解

x = arcsec (- 5 - 19), x = - arcsec (- 5 - 19) + 2 π

合并所有解

x = arcsec (- 5 - 19), x = - arcsec (- 5 - 19) + 2 π

以小数形式表示解

x = 1.67785 \dots, x = - 1.67785 \dots + 2 π

作图

流行的例子

tan (θ) = \frac{14}{11}

tan(3φ-4)=-1/2 ,0<= φ<= pi/2

tan (3 φ - 4) = - \frac{1}{2}, 0 \leq φ \leq \frac{π}{2}

solvefor g,θ=arctan((5.54)/(10g)+0.2)

so l v e f or g, θ = arctan (\frac{5.54}{10 g} + 0.2)

tan (t) = \frac{3}{4}

2cos(θ)-2sin(2θ)=0

2 cos (θ) - 2 sin (2 θ) = 0