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 >

cot^2(x)-3csc(x)+3=0

解答

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

解答

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{π}{2} + 2 πn

+1

度数

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n

求解步骤

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

使用三角恒等式改写

3 + cot^{2} (x) - 3 csc (x)

使用毕达哥拉斯恒等式:

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

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

= 3 + csc^{2} (x) - 1 - 3 csc (x)

化简

3 + csc^{2} (x) - 1 - 3 csc (x) : csc^{2} (x) - 3 csc (x) + 2

3 + csc^{2} (x) - 1 - 3 csc (x)

对同类项分组

= csc^{2} (x) - 3 csc (x) + 3 - 1

数字相加/相减:

3 - 1 = 2

= csc^{2} (x) - 3 csc (x) + 2

= csc^{2} (x) - 3 csc (x) + 2

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

用替代法求解

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

令

: csc (x) = u

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

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = 1, b = - 3, c = 2

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

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

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

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

使用指数法则:

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

若

n

是偶数

(- 3)^{2} = 3^{2}

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

数字相乘:

4 \cdot 1 \cdot 2 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

数字相减:

9 - 8 = 1

= 1

使用法则

1 = 1

= 1

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

将解分隔开

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

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

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

使用法则

- (- a) = a

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

数字相加:

3 + 1 = 4

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

数字相乘:

2 \cdot 1 = 2

= \frac{4}{2}

数字相除:

\frac{4}{2} = 2

= 2

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

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

使用法则

- (- a) = a

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

数字相减:

3 - 1 = 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 = 2, u = 1

u = csc (x)

代回

csc (x) = 2, csc (x) = 1

csc (x) = 2, csc (x) = 1

csc (x) = 2 : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

csc (x) = 2

csc (x) = 2

的通解

csc (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} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 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} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

csc (x) = 1 : x = \frac{π}{2} + 2 πn

csc (x) = 1

csc (x) = 1

的通解

csc (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} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 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} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

合并所有解

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{π}{2} + 2 πn

作图

流行的例子

2sin^2(θ/2-pi/6)-1=0

2 sin^{2} (\frac{θ}{2} - \frac{π}{6}) - 1 = 0

cos^2(36)-sin^2(36)=cos(x)

cos^{2} (3 6^{\circ}) - sin^{2} (3 6^{\circ}) = cos (x)

23cos(x)=-17-cos(x)

23 cos (x) = - 17 - cos (x)

8cos^2(θ)+14cos(θ)-11=8cos(θ)-6

8 cos^{2} (θ) + 14 cos (θ) - 11 = 8 cos (θ) - 6

sin^2(x)(tan^2(x))=1-sin^2(x)

sin^{2} (x) (tan^{2} (x)) = 1 - sin^{2} (x)