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 >

-8csc^2(x)-4cot(x)=-12

解答

- 8 csc^{2} (x) - 4 cot (x) = - 12

解答

x = \frac{3 π}{4} + πn, x = 1.10714 \dots + πn

+1

度数

x = 13 5^{\circ} + 18 0^{\circ} n, x = 63.43494 \dots^{\circ} + 18 0^{\circ} n

求解步骤

- 8 csc^{2} (x) - 4 cot (x) = - 12

两边减去

- 12

- 8 csc^{2} (x) - 4 cot (x) + 12 = 0

使用三角恒等式改写

12 - 4 cot (x) - 8 csc^{2} (x)

使用毕达哥拉斯恒等式:

csc^{2} (x) = 1 + cot^{2} (x)

= 12 - 4 cot (x) - 8 (1 + cot^{2} (x))

化简

12 - 4 cot (x) - 8 (1 + cot^{2} (x)) : - 8 cot^{2} (x) - 4 cot (x) + 4

12 - 4 cot (x) - 8 (1 + cot^{2} (x))

乘开

- 8 (1 + cot^{2} (x)) : - 8 - 8 cot^{2} (x)

- 8 (1 + cot^{2} (x))

使用分配律:

a (b + c) = ab + a c

a = - 8, b = 1, c = cot^{2} (x)

= - 8 \cdot 1 + (- 8) cot^{2} (x)

使用加减运算法则

+ (- a) = - a

= - 8 \cdot 1 - 8 cot^{2} (x)

数字相乘:

8 \cdot 1 = 8

= - 8 - 8 cot^{2} (x)

= 12 - 4 cot (x) - 8 - 8 cot^{2} (x)

化简

12 - 4 cot (x) - 8 - 8 cot^{2} (x) : - 8 cot^{2} (x) - 4 cot (x) + 4

12 - 4 cot (x) - 8 - 8 cot^{2} (x)

对同类项分组

= - 4 cot (x) - 8 cot^{2} (x) + 12 - 8

数字相加/相减:

12 - 8 = 4

= - 8 cot^{2} (x) - 4 cot (x) + 4

= - 8 cot^{2} (x) - 4 cot (x) + 4

= - 8 cot^{2} (x) - 4 cot (x) + 4

4 - 4 cot (x) - 8 cot^{2} (x) = 0

用替代法求解

4 - 4 cot (x) - 8 cot^{2} (x) = 0

令

: cot (x) = u

4 - 4 u - 8 u^{2} = 0

4 - 4 u - 8 u^{2} = 0 : u = - 1, u = \frac{1}{2}

4 - 4 u - 8 u^{2} = 0

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = - 8, b = - 4, c = 4

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

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

(- 4)^{2} - 4 (- 8) \cdot 4 = 12

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

使用法则

- (- a) = a

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

使用指数法则:

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

若

n

是偶数

(- 4)^{2} = 4^{2}

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

数字相乘:

4 \cdot 8 \cdot 4 = 128

= 4^{2} + 128

4^{2} = 16

= 16 + 128

数字相加:

16 + 128 = 144

= 144

因式分解数字:

144 = 1 2^{2}

= 1 2^{2}

使用根式运算法则:

n a^{n} = a

1 2^{2} = 12

= 12

u_{1, 2} = \frac{- ( - 4 ) \pm 12}{2 ( - 8 )}

将解分隔开

u_{1} = \frac{- ( - 4 ) + 12}{2 ( - 8 )}, u_{2} = \frac{- ( - 4 ) - 12}{2 ( - 8 )}

u = \frac{- ( - 4 ) + 12}{2 ( - 8 )} : - 1

\frac{- ( - 4 ) + 12}{2 ( - 8 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{4 + 12}{- 2 \cdot 8}

数字相加:

4 + 12 = 16

= \frac{16}{- 2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{16}{- 16}

使用分式法则:

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

= - \frac{16}{16}

使用法则

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 4 ) - 12}{2 ( - 8 )} : \frac{1}{2}

\frac{- ( - 4 ) - 12}{2 ( - 8 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{4 - 12}{- 2 \cdot 8}

数字相减:

4 - 12 = - 8

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

数字相乘:

2 \cdot 8 = 16

= \frac{- 8}{- 16}

使用分式法则:

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

= \frac{8}{16}

约分:

8

= \frac{1}{2}

二次方程组的解是:

u = - 1, u = \frac{1}{2}

u = cot (x)

代回

cot (x) = - 1, cot (x) = \frac{1}{2}

cot (x) = - 1, cot (x) = \frac{1}{2}

cot (x) = - 1 : x = \frac{3 π}{4} + πn

cot (x) = - 1

cot (x) = - 1

的通解

cot (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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

cot (x) = \frac{1}{2} : x = arccot (\frac{1}{2}) + πn

cot (x) = \frac{1}{2}

使用反三角函数性质

cot (x) = \frac{1}{2}

cot (x) = \frac{1}{2}

的通解

cot (x) = a \Rightarrow x = arccot (a) + πn

x = arccot (\frac{1}{2}) + πn

x = arccot (\frac{1}{2}) + πn

合并所有解

x = \frac{3 π}{4} + πn, x = arccot (\frac{1}{2}) + πn

以小数形式表示解

x = \frac{3 π}{4} + πn, x = 1.10714 \dots + πn

作图

流行的例子

sin(2x)+2cos^2(x)=0

sin (2 x) + 2 cos^{2} (x) = 0

sin (θ) = \frac{5}{3}

14sin^2(x)+5sin(x)-1=0

14 sin^{2} (x) + 5 sin (x) - 1 = 0

sin(x)= 3/5 sin(x+pi/2)+cos(pi-x)

sin (x) = \frac{3}{5} sin (x + \frac{π}{2}) + cos (π - x)

arccosh(x)=(2pi)/7

arccosh (x) = \frac{2 π}{7}