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 >

2cot^2(z)+5=7csc(z)

解答

2 cot^{2} (z) + 5 = 7 csc (z)

解答

z = 0.33983 \dots + 2 πn, z = π - 0.33983 \dots + 2 πn

+1

度数

z = 19.47122 \dots^{\circ} + 36 0^{\circ} n, z = 160.52877 \dots^{\circ} + 36 0^{\circ} n

求解步骤

2 cot^{2} (z) + 5 = 7 csc (z)

两边减去

7 csc (z)

2 cot^{2} (z) + 5 - 7 csc (z) = 0

使用三角恒等式改写

5 + 2 cot^{2} (z) - 7 csc (z)

使用毕达哥拉斯恒等式:

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

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

= 5 + 2 (csc^{2} (z) - 1) - 7 csc (z)

化简

5 + 2 (csc^{2} (z) - 1) - 7 csc (z) : 2 csc^{2} (z) - 7 csc (z) + 3

5 + 2 (csc^{2} (z) - 1) - 7 csc (z)

乘开

2 (csc^{2} (z) - 1) : 2 csc^{2} (z) - 2

2 (csc^{2} (z) - 1)

使用分配律:

a (b - c) = ab - a c

a = 2, b = csc^{2} (z), c = 1

= 2 csc^{2} (z) - 2 \cdot 1

数字相乘:

2 \cdot 1 = 2

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

= 5 + 2 csc^{2} (z) - 2 - 7 csc (z)

化简

5 + 2 csc^{2} (z) - 2 - 7 csc (z) : 2 csc^{2} (z) - 7 csc (z) + 3

5 + 2 csc^{2} (z) - 2 - 7 csc (z)

对同类项分组

= 2 csc^{2} (z) - 7 csc (z) + 5 - 2

数字相加/相减:

5 - 2 = 3

= 2 csc^{2} (z) - 7 csc (z) + 3

= 2 csc^{2} (z) - 7 csc (z) + 3

= 2 csc^{2} (z) - 7 csc (z) + 3

3 + 2 csc^{2} (z) - 7 csc (z) = 0

用替代法求解

3 + 2 csc^{2} (z) - 7 csc (z) = 0

令

: csc (z) = u

3 + 2 u^{2} - 7 u = 0

3 + 2 u^{2} - 7 u = 0 : u = 3, u = \frac{1}{2}

3 + 2 u^{2} - 7 u = 0

改写成标准形式

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

2 u^{2} - 7 u + 3 = 0

使用求根公式求解

2 u^{2} - 7 u + 3 = 0

二次方程求根公式:

若

a = 2, b = - 7, c = 3

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

(- 7)^{2} - 4 \cdot 2 \cdot 3 = 5

(- 7)^{2} - 4 \cdot 2 \cdot 3

使用指数法则:

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

若

n

是偶数

(- 7)^{2} = 7^{2}

= 7^{2} - 4 \cdot 2 \cdot 3

数字相乘:

4 \cdot 2 \cdot 3 = 24

= 7^{2} - 24

7^{2} = 49

= 49 - 24

数字相减:

49 - 24 = 25

= 25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 5

u_{1, 2} = \frac{- ( - 7 ) \pm 5}{2 \cdot 2}

将解分隔开

u_{1} = \frac{- ( - 7 ) + 5}{2 \cdot 2}, u_{2} = \frac{- ( - 7 ) - 5}{2 \cdot 2}

u = \frac{- ( - 7 ) + 5}{2 \cdot 2} : 3

\frac{- ( - 7 ) + 5}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{7 + 5}{2 \cdot 2}

数字相加:

7 + 5 = 12

= \frac{12}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{12}{4}

数字相除:

\frac{12}{4} = 3

= 3

u = \frac{- ( - 7 ) - 5}{2 \cdot 2} : \frac{1}{2}

\frac{- ( - 7 ) - 5}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{7 - 5}{2 \cdot 2}

数字相减:

7 - 5 = 2

= \frac{2}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{2}{4}

约分:

2

= \frac{1}{2}

二次方程组的解是:

u = 3, u = \frac{1}{2}

u = csc (z)

代回

csc (z) = 3, csc (z) = \frac{1}{2}

csc (z) = 3, csc (z) = \frac{1}{2}

csc (z) = 3 : z = arccsc (3) + 2 πn, z = π - arccsc (3) + 2 πn

csc (z) = 3

使用反三角函数性质

csc (z) = 3

csc (z) = 3

的通解

csc (x) = a \Rightarrow x = arccsc (a) + 2 πn, x = π - arccsc (a) + 2 πn

z = arccsc (3) + 2 πn, z = π - arccsc (3) + 2 πn

z = arccsc (3) + 2 πn, z = π - arccsc (3) + 2 πn

csc (z) = \frac{1}{2} :

无解

csc (z) = \frac{1}{2}

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

无解

合并所有解

z = arccsc (3) + 2 πn, z = π - arccsc (3) + 2 πn

以小数形式表示解

z = 0.33983 \dots + 2 πn, z = π - 0.33983 \dots + 2 πn

作图

流行的例子

sin^{2} (x) = 6

sin^2(φ)+cos^2(φ)=2

sin^{2} (φ) + cos^{2} (φ) = 2

solvefor x,y^2(2+sin(x))=1

so l v e f or x, y^{2} (2 + sin (x)) = 1

sqrt(2)=cos(x)+sin(x)

2 = cos (x) + sin (x)

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

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