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 >

2sin(θ)-3csc(θ)+5=0

解答

2 sin (θ) - 3 csc (θ) + 5 = 0

解答

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

+1

度数

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

求解步骤

2 sin (θ) - 3 csc (θ) + 5 = 0

使用三角恒等式改写

5 + 2 sin (θ) - 3 csc (θ)

使用基本三角恒等式:

sin (x) = \frac{1}{csc ( x )}

= 5 + 2 \cdot \frac{1}{csc ( θ )} - 3 csc (θ)

2 \cdot \frac{1}{csc ( θ )} = \frac{2}{csc ( θ )}

2 \cdot \frac{1}{csc ( θ )}

分式相乘:

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

= \frac{1 \cdot 2}{csc ( θ )}

数字相乘:

1 \cdot 2 = 2

= \frac{2}{csc ( θ )}

= 5 + \frac{2}{csc ( θ )} - 3 csc (θ)

5 + \frac{2}{csc ( θ )} - 3 csc (θ) = 0

用替代法求解

5 + \frac{2}{csc ( θ )} - 3 csc (θ) = 0

令

: csc (θ) = u

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

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

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

在两边乘以

u

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

在两边乘以

u

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

化简

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

化简

\frac{2}{u} u : 2

\frac{2}{u} u

分式相乘:

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

= \frac{2 u}{u}

约分:

u

= 2

化简

- 3 uu : - 3 u^{2}

- 3 uu

使用指数法则:

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

uu = u^{1 + 1}

= - 3 u^{1 + 1}

数字相加:

1 + 1 = 2

= - 3 u^{2}

化简

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

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

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

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

解

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

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

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

使用法则

- (- a) = a

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

数字相乘:

4 \cdot 3 \cdot 2 = 24

= 5^{2} + 24

5^{2} = 25

= 25 + 24

数字相加:

25 + 24 = 49

= 49

因式分解数字:

49 = 7^{2}

= 7^{2}

使用根式运算法则:

n a^{n} = a

7^{2} = 7

= 7

u_{1, 2} = \frac{- 5 \pm 7}{2 ( - 3 )}

将解分隔开

u_{1} = \frac{- 5 + 7}{2 ( - 3 )}, u_{2} = \frac{- 5 - 7}{2 ( - 3 )}

u = \frac{- 5 + 7}{2 ( - 3 )} : - \frac{1}{3}

\frac{- 5 + 7}{2 ( - 3 )}

去除括号:

(- a) = - a

= \frac{- 5 + 7}{- 2 \cdot 3}

数字相加/相减:

- 5 + 7 = 2

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

数字相乘:

2 \cdot 3 = 6

= \frac{2}{- 6}

使用分式法则:

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

= - \frac{2}{6}

约分:

2

= - \frac{1}{3}

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

\frac{- 5 - 7}{2 ( - 3 )}

去除括号:

(- a) = - a

= \frac{- 5 - 7}{- 2 \cdot 3}

数字相减:

- 5 - 7 = - 12

= \frac{- 12}{- 2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{- 12}{- 6}

使用分式法则:

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

= \frac{12}{6}

数字相除:

\frac{12}{6} = 2

= 2

二次方程组的解是:

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

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

验证解

找到无定义的点（奇点）:

u = 0

取

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

u = csc (θ)

代回

csc (θ) = - \frac{1}{3}, csc (θ) = 2

csc (θ) = - \frac{1}{3}, csc (θ) = 2

csc (θ) = - \frac{1}{3} :

无解

csc (θ) = - \frac{1}{3}

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

无解

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

csc (θ) = 2

csc (θ) = 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

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

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

合并所有解

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

作图

流行的例子

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

sin (θ) = \frac{4}{13}

sin(3x)=-1/2 ,0<= x<= 2pi

sin (3 x) = - \frac{1}{2}, 0 \leq x \leq 2 π

4cos(x)-3sec(x)=0

4 cos (x) - 3 sec (x) = 0

sin (θ) = \frac{11}{61}