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 >

sec(x)+4cos(x)=5

解答

sec (x) + 4 cos (x) = 5

解答

x = 1.31811 \dots + 2 πn, x = 2 π - 1.31811 \dots + 2 πn, x = 2 πn

+1

度数

x = 75.52248 \dots^{\circ} + 36 0^{\circ} n, x = 284.47751 \dots^{\circ} + 36 0^{\circ} n, x = 0^{\circ} + 36 0^{\circ} n

求解步骤

sec (x) + 4 cos (x) = 5

两边减去

5

sec (x) + 4 cos (x) - 5 = 0

使用三角恒等式改写

- 5 + sec (x) + 4 cos (x)

使用基本三角恒等式:

cos (x) = \frac{1}{sec ( x )}

= - 5 + sec (x) + 4 \cdot \frac{1}{sec ( x )}

4 \cdot \frac{1}{sec ( x )} = \frac{4}{sec ( x )}

4 \cdot \frac{1}{sec ( x )}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 4}{sec ( x )}

数字相乘:

1 \cdot 4 = 4

= \frac{4}{sec ( x )}

= - 5 + sec (x) + \frac{4}{sec ( x )}

- 5 + \frac{4}{sec ( x )} + sec (x) = 0

用替代法求解

- 5 + \frac{4}{sec ( x )} + sec (x) = 0

令

: sec (x) = u

- 5 + \frac{4}{u} + u = 0

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

- 5 + \frac{4}{u} + u = 0

在两边乘以

u

- 5 + \frac{4}{u} + u = 0

在两边乘以

u

- 5 u + \frac{4}{u} u + uu = 0 \cdot u

化简

- 5 u + \frac{4}{u} u + uu = 0 \cdot u

化简

\frac{4}{u} u : 4

\frac{4}{u} u

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{4 u}{u}

约分:

u

= 4

化简

uu : u^{2}

uu

使用指数法则:

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

uu = u^{1 + 1}

= u^{1 + 1}

数字相加:

1 + 1 = 2

= u^{2}

化简

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

- 5 u + 4 + u^{2} = 0

- 5 u + 4 + u^{2} = 0

- 5 u + 4 + u^{2} = 0

解

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

- 5 u + 4 + u^{2} = 0

改写成标准形式

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

u^{2} - 5 u + 4 = 0

使用求根公式求解

u^{2} - 5 u + 4 = 0

二次方程求根公式:

若

a = 1, b = - 5, c = 4

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

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

(- 5)^{2} - 4 \cdot 1 \cdot 4 = 3

(- 5)^{2} - 4 \cdot 1 \cdot 4

使用指数法则:

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

若

n

是偶数

(- 5)^{2} = 5^{2}

= 5^{2} - 4 \cdot 1 \cdot 4

数字相乘:

4 \cdot 1 \cdot 4 = 16

= 5^{2} - 16

5^{2} = 25

= 25 - 16

数字相减:

25 - 16 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 3

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

将解分隔开

u_{1} = \frac{- ( - 5 ) + 3}{2 \cdot 1}, u_{2} = \frac{- ( - 5 ) - 3}{2 \cdot 1}

u = \frac{- ( - 5 ) + 3}{2 \cdot 1} : 4

\frac{- ( - 5 ) + 3}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{5 + 3}{2 \cdot 1}

数字相加:

5 + 3 = 8

= \frac{8}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{8}{2}

数字相除:

\frac{8}{2} = 4

= 4

u = \frac{- ( - 5 ) - 3}{2 \cdot 1} : 1

\frac{- ( - 5 ) - 3}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{5 - 3}{2 \cdot 1}

数字相减:

5 - 3 = 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

二次方程组的解是:

u = 4, u = 1

u = 4, u = 1

验证解

找到无定义的点（奇点）:

u = 0

取

- 5 + \frac{4}{u} + u

的分母，令其等于零

u = 0

以下点无定义

u = 0

将不在定义域的点与解相综合:

u = 4, u = 1

u = sec (x)

代回

sec (x) = 4, sec (x) = 1

sec (x) = 4, sec (x) = 1

sec (x) = 4 : x = arcsec (4) + 2 πn, x = 2 π - arcsec (4) + 2 πn

sec (x) = 4

使用反三角函数性质

sec (x) = 4

sec (x) = 4

的通解

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

x = arcsec (4) + 2 πn, x = 2 π - arcsec (4) + 2 πn

x = arcsec (4) + 2 πn, x = 2 π - arcsec (4) + 2 πn

sec (x) = 1 : x = 2 πn

sec (x) = 1

sec (x) = 1

的通解

sec (x)

周期表（周期为

2 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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

x = 0 + 2 πn

x = 0 + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

合并所有解

x = arcsec (4) + 2 πn, x = 2 π - arcsec (4) + 2 πn, x = 2 πn

以小数形式表示解

x = 1.31811 \dots + 2 πn, x = 2 π - 1.31811 \dots + 2 πn, x = 2 πn

作图

流行的例子

-sqrt(3)sec(x)=2

- 3 sec (x) = 2

3cos(x)csc(x)=2sqrt(3)cos(x)

3 cos (x) csc (x) = 23 cos (x)

cos (2 x) = \frac{9}{41}

sec(5x)= 2/(sqrt(3))

sec (5 x) = \frac{2}{3}

cos(a)=-4/9 , pi/2 <= a<= pi

cos (a) = - \frac{4}{9}, \frac{π}{2} \leq a \leq π