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受欢迎的三角函数 >

6sin(θ)-3csc(θ)=0

解答

6 sin (θ) - 3 csc (θ) = 0

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

6 sin (θ) - 3 csc (θ) = 0

使用三角恒等式改写

- 3 csc (θ) + 6 sin (θ)

使用基本三角恒等式:

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

= - 3 csc (θ) + 6 \cdot \frac{1}{csc ( θ )}

6 \cdot \frac{1}{csc ( θ )} = \frac{6}{csc ( θ )}

6 \cdot \frac{1}{csc ( θ )}

分式相乘:

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

= \frac{1 \cdot 6}{csc ( θ )}

数字相乘:

1 \cdot 6 = 6

= \frac{6}{csc ( θ )}

= - 3 csc (θ) + \frac{6}{csc ( θ )}

\frac{6}{csc ( θ )} - 3 csc (θ) = 0

用替代法求解

\frac{6}{csc ( θ )} - 3 csc (θ) = 0

令

: csc (θ) = u

\frac{6}{u} - 3 u = 0

\frac{6}{u} - 3 u = 0 : u = 2, u = - 2

\frac{6}{u} - 3 u = 0

在两边乘以

u

\frac{6}{u} - 3 u = 0

在两边乘以

u

\frac{6}{u} u - 3 uu = 0 \cdot u

化简

\frac{6}{u} u - 3 uu = 0 \cdot u

化简

\frac{6}{u} u : 6

\frac{6}{u} u

分式相乘:

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

= \frac{6 u}{u}

约分:

u

= 6

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

6 - 3 u^{2} = 0

6 - 3 u^{2} = 0

6 - 3 u^{2} = 0

解

6 - 3 u^{2} = 0 : u = 2, u = - 2

6 - 3 u^{2} = 0

将

6

到右边

6 - 3 u^{2} = 0

两边减去

6

6 - 3 u^{2} - 6 = 0 - 6

化简

- 3 u^{2} = - 6

- 3 u^{2} = - 6

两边除以

- 3

- 3 u^{2} = - 6

两边除以

- 3

\frac{- 3 u ^{2}}{- 3} = \frac{- 6}{- 3}

化简

\frac{- 3 u ^{2}}{- 3} = \frac{- 6}{- 3}

化简

\frac{- 3 u ^{2}}{- 3} : u^{2}

\frac{- 3 u ^{2}}{- 3}

使用分式法则:

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

= \frac{3 u ^{2}}{3}

数字相除:

\frac{3}{3} = 1

= u^{2}

化简

\frac{- 6}{- 3} : 2

\frac{- 6}{- 3}

使用分式法则:

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

= \frac{6}{3}

数字相除:

\frac{6}{3} = 2

= 2

u^{2} = 2

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

\frac{6}{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 = 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

tan (x) = 1

cos (7 x) = 0

cos (7 x) = 1

cos (x + \frac{π}{2}) = 0

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