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受欢迎的三角函数 >

8sin(x)-4csc(x)=0

解答

8 sin (x) - 4 csc (x) = 0

解答

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

+1

度数

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

求解步骤

8 sin (x) - 4 csc (x) = 0

使用三角恒等式改写

- 4 csc (x) + 8 sin (x)

使用基本三角恒等式:

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

= - 4 csc (x) + 8 \cdot \frac{1}{csc ( x )}

8 \cdot \frac{1}{csc ( x )} = \frac{8}{csc ( x )}

8 \cdot \frac{1}{csc ( x )}

分式相乘:

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

= \frac{1 \cdot 8}{csc ( x )}

数字相乘:

1 \cdot 8 = 8

= \frac{8}{csc ( x )}

= - 4 csc (x) + \frac{8}{csc ( x )}

\frac{8}{csc ( x )} - 4 csc (x) = 0

用替代法求解

\frac{8}{csc ( x )} - 4 csc (x) = 0

令

: csc (x) = u

\frac{8}{u} - 4 u = 0

\frac{8}{u} - 4 u = 0 : u = 2, u = - 2

\frac{8}{u} - 4 u = 0

在两边乘以

u

\frac{8}{u} - 4 u = 0

在两边乘以

u

\frac{8}{u} u - 4 uu = 0 \cdot u

化简

\frac{8}{u} u - 4 uu = 0 \cdot u

化简

\frac{8}{u} u : 8

\frac{8}{u} u

分式相乘:

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

= \frac{8 u}{u}

约分:

u

= 8

化简

- 4 uu : - 4 u^{2}

- 4 uu

使用指数法则:

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

uu = u^{1 + 1}

= - 4 u^{1 + 1}

数字相加:

1 + 1 = 2

= - 4 u^{2}

化简

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

8 - 4 u^{2} = 0

8 - 4 u^{2} = 0

8 - 4 u^{2} = 0

解

8 - 4 u^{2} = 0 : u = 2, u = - 2

8 - 4 u^{2} = 0

将

8

到右边

8 - 4 u^{2} = 0

两边减去

8

8 - 4 u^{2} - 8 = 0 - 8

化简

- 4 u^{2} = - 8

- 4 u^{2} = - 8

两边除以

- 4

- 4 u^{2} = - 8

两边除以

- 4

\frac{- 4 u ^{2}}{- 4} = \frac{- 8}{- 4}

化简

\frac{- 4 u ^{2}}{- 4} = \frac{- 8}{- 4}

化简

\frac{- 4 u ^{2}}{- 4} : u^{2}

\frac{- 4 u ^{2}}{- 4}

使用分式法则:

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

= \frac{4 u ^{2}}{4}

数字相除:

\frac{4}{4} = 1

= u^{2}

化简

\frac{- 8}{- 4} : 2

\frac{- 8}{- 4}

使用分式法则:

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

= \frac{8}{4}

数字相除:

\frac{8}{4} = 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{8}{u} - 4 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 (x)

代回

csc (x) = 2, csc (x) = - 2

csc (x) = 2, csc (x) = - 2

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

csc (x) = 2

csc (x) = 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

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

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

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

csc (x) = - 2

csc (x) = - 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

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

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

合并所有解

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

作图

流行的例子

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