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受欢迎的三角函数 >

3+sin(θ)=2csc(θ)

解答

3 + sin (θ) = 2 csc (θ)

解答

θ = 0.59626 \dots + 2 πn, θ = π - 0.59626 \dots + 2 πn

+1

度数

θ = 34.16325 \dots^{\circ} + 36 0^{\circ} n, θ = 145.83674 \dots^{\circ} + 36 0^{\circ} n

求解步骤

3 + sin (θ) = 2 csc (θ)

两边减去

2 csc (θ)

3 + sin (θ) - 2 csc (θ) = 0

使用三角恒等式改写

3 + sin (θ) - 2 csc (θ)

使用基本三角恒等式:

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

= 3 + \frac{1}{csc ( θ )} - 2 csc (θ)

3 + \frac{1}{csc ( θ )} - 2 csc (θ) = 0

用替代法求解

3 + \frac{1}{csc ( θ )} - 2 csc (θ) = 0

令

: csc (θ) = u

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

3 + \frac{1}{u} - 2 u = 0 : u = - \frac{- 3 + 17}{4}, u = \frac{3 + 17}{4}

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

在两边乘以

u

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

在两边乘以

u

3 u + \frac{1}{u} u - 2 uu = 0 \cdot u

化简

3 u + \frac{1}{u} u - 2 uu = 0 \cdot u

化简

\frac{1}{u} u : 1

\frac{1}{u} u

分式相乘:

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

= \frac{1 \cdot u}{u}

约分:

u

= 1

化简

- 2 uu : - 2 u^{2}

- 2 uu

使用指数法则:

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

uu = u^{1 + 1}

= - 2 u^{1 + 1}

数字相加:

1 + 1 = 2

= - 2 u^{2}

化简

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

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

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

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

解

3 u + 1 - 2 u^{2} = 0 : u = - \frac{- 3 + 17}{4}, u = \frac{3 + 17}{4}

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

3^{2} - 4 (- 2) \cdot 1 = 17

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

使用法则

- (- a) = a

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

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 3^{2} + 8

3^{2} = 9

= 9 + 8

数字相加:

9 + 8 = 17

= 17

u_{1, 2} = \frac{- 3 \pm 17}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- 3 + 17}{2 ( - 2 )}, u_{2} = \frac{- 3 - 17}{2 ( - 2 )}

u = \frac{- 3 + 17}{2 ( - 2 )} : - \frac{- 3 + 17}{4}

\frac{- 3 + 17}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 3 + 17}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 3 + 17}{- 4}

使用分式法则:

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

= - \frac{- 3 + 17}{4}

u = \frac{- 3 - 17}{2 ( - 2 )} : \frac{3 + 17}{4}

\frac{- 3 - 17}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 3 - 17}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 3 - 17}{- 4}

使用分式法则:

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

- 3 - 17 = - (3 + 17)

= \frac{3 + 17}{4}

二次方程组的解是:

u = - \frac{- 3 + 17}{4}, u = \frac{3 + 17}{4}

u = - \frac{- 3 + 17}{4}, u = \frac{3 + 17}{4}

验证解

找到无定义的点（奇点）:

u = 0

取

3 + \frac{1}{u} - 2 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{- 3 + 17}{4}, u = \frac{3 + 17}{4}

u = csc (θ)

代回

csc (θ) = - \frac{- 3 + 17}{4}, csc (θ) = \frac{3 + 17}{4}

csc (θ) = - \frac{- 3 + 17}{4}, csc (θ) = \frac{3 + 17}{4}

csc (θ) = - \frac{- 3 + 17}{4} :

无解

csc (θ) = - \frac{- 3 + 17}{4}

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

无解

csc (θ) = \frac{3 + 17}{4} : θ = arccsc (\frac{3 + 17}{4}) + 2 πn, θ = π - arccsc (\frac{3 + 17}{4}) + 2 πn

csc (θ) = \frac{3 + 17}{4}

使用反三角函数性质

csc (θ) = \frac{3 + 17}{4}

csc (θ) = \frac{3 + 17}{4}

的通解

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

θ = arccsc (\frac{3 + 17}{4}) + 2 πn, θ = π - arccsc (\frac{3 + 17}{4}) + 2 πn

θ = arccsc (\frac{3 + 17}{4}) + 2 πn, θ = π - arccsc (\frac{3 + 17}{4}) + 2 πn

合并所有解

θ = arccsc (\frac{3 + 17}{4}) + 2 πn, θ = π - arccsc (\frac{3 + 17}{4}) + 2 πn

以小数形式表示解

θ = 0.59626 \dots + 2 πn, θ = π - 0.59626 \dots + 2 πn

作图

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