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 >

5sec(x)-2=3cos(x)

解答

5 sec (x) - 2 = 3 cos (x)

解答

x = 2 πn

+1

度数

x = 0^{\circ} + 36 0^{\circ} n

求解步骤

5 sec (x) - 2 = 3 cos (x)

两边减去

3 cos (x)

5 sec (x) - 2 - 3 cos (x) = 0

使用三角恒等式改写

- 2 - 3 cos (x) + 5 sec (x)

使用基本三角恒等式:

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

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

3 \cdot \frac{1}{sec ( x )} = \frac{3}{sec ( x )}

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

分式相乘:

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

= \frac{1 \cdot 3}{sec ( x )}

数字相乘:

1 \cdot 3 = 3

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

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

- 2 - \frac{3}{sec ( x )} + 5 sec (x) = 0

用替代法求解

- 2 - \frac{3}{sec ( x )} + 5 sec (x) = 0

令

: sec (x) = u

- 2 - \frac{3}{u} + 5 u = 0

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

- 2 - \frac{3}{u} + 5 u = 0

在两边乘以

u

- 2 - \frac{3}{u} + 5 u = 0

在两边乘以

u

- 2 u - \frac{3}{u} u + 5 uu = 0 \cdot u

化简

- 2 u - \frac{3}{u} u + 5 uu = 0 \cdot u

化简

- \frac{3}{u} u : - 3

- \frac{3}{u} u

分式相乘:

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

= - \frac{3 u}{u}

约分:

u

= - 3

化简

5 uu : 5 u^{2}

5 uu

使用指数法则:

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

uu = u^{1 + 1}

= 5 u^{1 + 1}

数字相加:

1 + 1 = 2

= 5 u^{2}

化简

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

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

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

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

解

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = 5, b = - 2, c = - 3

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

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

(- 2)^{2} - 4 \cdot 5 (- 3) = 8

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

使用法则

- (- a) = a

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

使用指数法则:

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

若

n

是偶数

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

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

数字相乘:

4 \cdot 5 \cdot 3 = 60

= 2^{2} + 60

2^{2} = 4

= 4 + 60

数字相加:

4 + 60 = 64

= 64

因式分解数字:

64 = 8^{2}

= 8^{2}

使用根式运算法则:

n a^{n} = a

8^{2} = 8

= 8

u_{1, 2} = \frac{- ( - 2 ) \pm 8}{2 \cdot 5}

将解分隔开

u_{1} = \frac{- ( - 2 ) + 8}{2 \cdot 5}, u_{2} = \frac{- ( - 2 ) - 8}{2 \cdot 5}

u = \frac{- ( - 2 ) + 8}{2 \cdot 5} : 1

\frac{- ( - 2 ) + 8}{2 \cdot 5}

使用法则

- (- a) = a

= \frac{2 + 8}{2 \cdot 5}

数字相加:

2 + 8 = 10

= \frac{10}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{10}{10}

使用法则

\frac{a}{a} = 1

= 1

u = \frac{- ( - 2 ) - 8}{2 \cdot 5} : - \frac{3}{5}

\frac{- ( - 2 ) - 8}{2 \cdot 5}

使用法则

- (- a) = a

= \frac{2 - 8}{2 \cdot 5}

数字相减:

2 - 8 = - 6

= \frac{- 6}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{- 6}{10}

使用分式法则:

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

= - \frac{6}{10}

约分:

2

= - \frac{3}{5}

二次方程组的解是:

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

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

验证解

找到无定义的点（奇点）:

u = 0

取

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

u = sec (x)

代回

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

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

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

sec (x) = - \frac{3}{5} :

无解

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

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

无解

合并所有解

x = 2 πn

作图

流行的例子

(sin(x))(cos(x))(tan(x))+cos(2x)=1

(sin (x)) (cos (x)) (tan (x)) + cos (2 x) = 1

tan (x) = \frac{20}{30}

sin (x) = sin (2 7^{\circ})

cos (a) = \frac{12}{37}

120sin(x-120)=120sin(x-240)

120 sin (x - 12 0^{\circ}) = 120 sin (x - 24 0^{\circ})