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受欢迎的三角函数 >

2sin(θ)+3=-csc(θ)

解答

2 sin (θ) + 3 = - csc (θ)

解答

θ = \frac{3 π}{2} + 2 πn, θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

+1

度数

θ = 27 0^{\circ} + 36 0^{\circ} n, θ = 21 0^{\circ} + 36 0^{\circ} n, θ = 33 0^{\circ} + 36 0^{\circ} n

求解步骤

2 sin (θ) + 3 = - csc (θ)

两边减去

- csc (θ)

2 sin (θ) + 3 + csc (θ) = 0

使用三角恒等式改写

3 + csc (θ) + 2 sin (θ)

使用基本三角恒等式:

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

= 3 + csc (θ) + 2 \cdot \frac{1}{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 )}

= 3 + csc (θ) + \frac{2}{csc ( θ )}

3 + csc (θ) + \frac{2}{csc ( θ )} = 0

用替代法求解

3 + csc (θ) + \frac{2}{csc ( θ )} = 0

令

: csc (θ) = u

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

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

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

在两边乘以

u

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

在两边乘以

u

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

化简

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

化简

uu : u^{2}

uu

使用指数法则:

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

uu = u^{1 + 1}

= u^{1 + 1}

数字相加:

1 + 1 = 2

= u^{2}

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

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

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

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

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

解

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

3 u + u^{2} + 2 = 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

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} : - 1

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

数字相加/相减:

- 3 + 1 = - 2

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

数字相乘:

2 \cdot 1 = 2

= \frac{- 2}{2}

使用分式法则:

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

= - \frac{2}{2}

使用法则

\frac{a}{a} = 1

= - 1

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

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

数字相减:

- 3 - 1 = - 4

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

数字相乘:

2 \cdot 1 = 2

= \frac{- 4}{2}

使用分式法则:

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

= - \frac{4}{2}

数字相除:

\frac{4}{2} = 2

= - 2

二次方程组的解是:

u = - 1, u = - 2

u = - 1, u = - 2

验证解

找到无定义的点（奇点）:

u = 0

取

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

u = csc (θ)

代回

csc (θ) = - 1, csc (θ) = - 2

csc (θ) = - 1, csc (θ) = - 2

csc (θ) = - 1 : θ = \frac{3 π}{2} + 2 πn

csc (θ) = - 1

csc (θ) = - 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

θ = \frac{3 π}{2} + 2 πn

θ = \frac{3 π}{2} + 2 πn

csc (θ) = - 2 : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 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{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

合并所有解

θ = \frac{3 π}{2} + 2 πn, θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

作图

流行的例子

2sin((2x)/3)-sqrt(2)=0

2 sin (\frac{2 x}{3}) - 2 = 0

tan (θ) = - \frac{1}{4}

3-sin(θ)=cos(2θ)

3 - sin (θ) = cos (2 θ)

tan (θ) = \frac{- 1}{1}

cot (x) = 3