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受欢迎的三角函数 >

4sin(2x)+csc(2x)=4

解答

4 sin (2 x) + csc (2 x) = 4

解答

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

+1

度数

x = 1 5^{\circ} + 18 0^{\circ} n, x = 7 5^{\circ} + 18 0^{\circ} n

求解步骤

4 sin (2 x) + csc (2 x) = 4

两边减去

4

4 sin (2 x) + csc (2 x) - 4 = 0

使用三角恒等式改写

- 4 + csc (2 x) + 4 sin (2 x)

使用基本三角恒等式:

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

= - 4 + csc (2 x) + 4 \cdot \frac{1}{csc ( 2 x )}

4 \cdot \frac{1}{csc ( 2 x )} = \frac{4}{csc ( 2 x )}

4 \cdot \frac{1}{csc ( 2 x )}

分式相乘:

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

= \frac{1 \cdot 4}{csc ( 2 x )}

数字相乘:

1 \cdot 4 = 4

= \frac{4}{csc ( 2 x )}

= - 4 + csc (2 x) + \frac{4}{csc ( 2 x )}

- 4 + csc (2 x) + \frac{4}{csc ( 2 x )} = 0

用替代法求解

- 4 + csc (2 x) + \frac{4}{csc ( 2 x )} = 0

令

: csc (2 x) = u

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

- 4 + u + \frac{4}{u} = 0 : u = 2

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

在两边乘以

u

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

在两边乘以

u

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

化简

- 4 u + uu + \frac{4}{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{4}{u} u : 4

\frac{4}{u} u

分式相乘:

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

= \frac{4 u}{u}

约分:

u

= 4

化简

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

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

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

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

解

- 4 u + u^{2} + 4 = 0 : u = 2

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = 1, b = - 4, c = 4

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

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

(- 4)^{2} - 4 \cdot 1 \cdot 4 = 0

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

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 4)^{2} = 4^{2}

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

数字相乘:

4 \cdot 1 \cdot 4 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

数字相减:

16 - 16 = 0

= 0

u_{1, 2} = \frac{- ( - 4 ) \pm 0}{2 \cdot 1}

u = \frac{- ( - 4 )}{2 \cdot 1}

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

\frac{- ( - 4 )}{2 \cdot 1}

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 1 = 2

= \frac{4}{2}

数字相除:

\frac{4}{2} = 2

= 2

u = 2

二次方程组的解是:

u = 2

u = 2

验证解

找到无定义的点（奇点）:

u = 0

取

- 4 + u + \frac{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 = csc (2 x)

代回

csc (2 x) = 2

csc (2 x) = 2

csc (2 x) = 2 : x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

csc (2 x) = 2

csc (2 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

2 x = \frac{π}{6} + 2 πn, 2 x = \frac{5 π}{6} + 2 πn

2 x = \frac{π}{6} + 2 πn, 2 x = \frac{5 π}{6} + 2 πn

解

2 x = \frac{π}{6} + 2 πn : x = \frac{π}{12} + πn

2 x = \frac{π}{6} + 2 πn

两边除以

2

2 x = \frac{π}{6} + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} = \frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{\frac{π}{6}}{2} + \frac{2 πn}{2} : \frac{π}{12} + πn

\frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{6}}{2} = \frac{π}{12}

\frac{\frac{π}{6}}{2}

使用分式法则:

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

= \frac{π}{6 \cdot 2}

数字相乘:

6 \cdot 2 = 12

= \frac{π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{π}{12} + πn

x = \frac{π}{12} + πn

x = \frac{π}{12} + πn

x = \frac{π}{12} + πn

解

2 x = \frac{5 π}{6} + 2 πn : x = \frac{5 π}{12} + πn

2 x = \frac{5 π}{6} + 2 πn

两边除以

2

2 x = \frac{5 π}{6} + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} = \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2} : \frac{5 π}{12} + πn

\frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{6}}{2} = \frac{5 π}{12}

\frac{\frac{5 π}{6}}{2}

使用分式法则:

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

= \frac{5 π}{6 \cdot 2}

数字相乘:

6 \cdot 2 = 12

= \frac{5 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

合并所有解

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

作图

流行的例子

cot (θ) = \frac{9}{40}

sqrt(5)cos(x)-2sin(x)=0.5

5 cos (x) - 2 sin (x) = 0.5

sin (x) = \frac{6}{15}

cot (θ) = 1.4

75/125 =(cosh(0.2x))/(cosh(0.4x))

\frac{75}{125} = \frac{cosh ( 0.2 x )}{cosh ( 0.4 x )}