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受欢迎的三角函数 >

2sin(θ)=csc(θ)

解答

2 sin (θ) = csc (θ)

解答

θ = \frac{π}{4} + 2 πn, θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

+1

度数

θ = 4 5^{\circ} + 36 0^{\circ} n, θ = 13 5^{\circ} + 36 0^{\circ} n, θ = 22 5^{\circ} + 36 0^{\circ} n, θ = 31 5^{\circ} + 36 0^{\circ} n

求解步骤

2 sin (θ) = csc (θ)

两边减去

csc (θ)

2 sin (θ) - csc (θ) = 0

使用三角恒等式改写

- csc (θ) + 2 sin (θ)

使用基本三角恒等式:

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

= - 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 )}

= - csc (θ) + \frac{2}{csc ( θ )}

- csc (θ) + \frac{2}{csc ( θ )} = 0

用替代法求解

- csc (θ) + \frac{2}{csc ( θ )} = 0

令

: csc (θ) = u

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

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

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

在两边乘以

u

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

在两边乘以

u

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

化简

- 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

- u^{2} + 2 = 0

- u^{2} + 2 = 0

- u^{2} + 2 = 0

解

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

- u^{2} + 2 = 0

将

2

到右边

- u^{2} + 2 = 0

两边减去

2

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

化简

- u^{2} = - 2

- u^{2} = - 2

两边除以

- 1

- u^{2} = - 2

两边除以

- 1

\frac{- u ^{2}}{- 1} = \frac{- 2}{- 1}

化简

u^{2} = 2

u^{2} = 2

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = 2, u = - 2

u = 2, u = - 2

验证解

找到无定义的点（奇点）:

u = 0

取

- 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 = 2, u = - 2

u = csc (θ)

代回

csc (θ) = 2, csc (θ) = - 2

csc (θ) = 2, csc (θ) = - 2

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

θ = \frac{π}{4} + 2 πn, θ = \frac{3 π}{4} + 2 πn

csc (θ) = - 2 : θ = \frac{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 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{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

θ = \frac{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

合并所有解

θ = \frac{π}{4} + 2 πn, θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

作图

流行的例子

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