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 >

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

解答

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

解答

x = 0.32175 \dots + πn, x = \frac{π}{4} + πn

+1

度数

x = 18.43494 \dots^{\circ} + 18 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

求解步骤

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

两边减去

- 2

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

使用三角恒等式改写

2 + csc^{2} (x) - 4 cot (x)

使用毕达哥拉斯恒等式:

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

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

化简

= cot^{2} (x) - 4 cot (x) + 3

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

用替代法求解

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

令

: cot (x) = u

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

3 + u^{2} - 4 u = 0 : u = 3, u = 1

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

改写成标准形式

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

u^{2} - 4 u + 3 = 0

使用求根公式求解

u^{2} - 4 u + 3 = 0

二次方程求根公式:

若

a = 1, b = - 4, c = 3

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

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

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

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

使用指数法则:

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

若

n

是偶数

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

= 4^{2} - 4 \cdot 1 \cdot 3

数字相乘:

4 \cdot 1 \cdot 3 = 12

= 4^{2} - 12

4^{2} = 16

= 16 - 12

数字相减:

16 - 12 = 4

= 4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

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

将解分隔开

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

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

\frac{- ( - 4 ) + 2}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{4 + 2}{2 \cdot 1}

数字相加:

4 + 2 = 6

= \frac{6}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{6}{2}

数字相除:

\frac{6}{2} = 3

= 3

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

\frac{- ( - 4 ) - 2}{2 \cdot 1}

使用法则

- (- a) = a

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

数字相减:

4 - 2 = 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 = 3, u = 1

u = cot (x)

代回

cot (x) = 3, cot (x) = 1

cot (x) = 3, cot (x) = 1

cot (x) = 3 : x = arccot (3) + πn

cot (x) = 3

使用反三角函数性质

cot (x) = 3

cot (x) = 3

的通解

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

x = arccot (3) + πn

x = arccot (3) + πn

cot (x) = 1 : x = \frac{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{π}{4} + πn

x = \frac{π}{4} + πn

合并所有解

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

以小数形式表示解

x = 0.32175 \dots + πn, x = \frac{π}{4} + πn

作图

流行的例子

sin(2x-pi/3)=(sqrt(3))/2

sin (2 x - \frac{π}{3}) = \frac{3}{2}

-sin^3(t)+sin(2t)cos(t)=0

- sin^{3} (t) + sin (2 t) cos (t) = 0

tan(θ)=sqrt(3)tan(45)

tan (θ) = 3 tan (4 5^{\circ})

2cos(2θ)-3cos(θ)-5=-7

2 cos (2 θ) - 3 cos (θ) - 5 = - 7

cos(x)tan(x)=-1/2

cos (x) tan (x) = - \frac{1}{2}