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受欢迎的三角函数 >

3sin(x)+1=2csc(x)

解答

3 sin (x) + 1 = 2 csc (x)

解答

x = \frac{3 π}{2} + 2 πn, x = 0.72972 \dots + 2 πn, x = π - 0.72972 \dots + 2 πn

+1

度数

x = 27 0^{\circ} + 36 0^{\circ} n, x = 41.81031 \dots^{\circ} + 36 0^{\circ} n, x = 138.18968 \dots^{\circ} + 36 0^{\circ} n

求解步骤

3 sin (x) + 1 = 2 csc (x)

两边减去

2 csc (x)

3 sin (x) + 1 - 2 csc (x) = 0

使用三角恒等式改写

1 - 2 csc (x) + 3 sin (x)

使用基本三角恒等式:

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

= 1 - 2 csc (x) + 3 \cdot \frac{1}{csc ( x )}

3 \cdot \frac{1}{csc ( x )} = \frac{3}{csc ( x )}

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

分式相乘:

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

= \frac{1 \cdot 3}{csc ( x )}

数字相乘:

1 \cdot 3 = 3

= \frac{3}{csc ( x )}

= 1 - 2 csc (x) + \frac{3}{csc ( x )}

1 + \frac{3}{csc ( x )} - 2 csc (x) = 0

用替代法求解

1 + \frac{3}{csc ( x )} - 2 csc (x) = 0

令

: csc (x) = u

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

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

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

在两边乘以

u

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

在两边乘以

u

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

化简

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

化简

1 \cdot u : u

1 \cdot u

乘以:

1 \cdot u = u

= u

化简

\frac{3}{u} u : 3

\frac{3}{u} u

分式相乘:

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

= \frac{3 u}{u}

约分:

u

= 3

化简

- 2 uu : - 2 u^{2}

- 2 uu

使用指数法则:

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

uu = u^{1 + 1}

= - 2 u^{1 + 1}

数字相加:

1 + 1 = 2

= - 2 u^{2}

化简

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

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

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

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

解

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = - 2, b = 1, c = 3

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 2 ) \cdot 3}{2 ( - 2 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 2 ) \cdot 3}{2 ( - 2 )}

1^{2} - 4 (- 2) \cdot 3 = 5

1^{2} - 4 (- 2) \cdot 3

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 2) \cdot 3

使用法则

- (- a) = a

= 1 + 4 \cdot 2 \cdot 3

数字相乘:

4 \cdot 2 \cdot 3 = 24

= 1 + 24

数字相加:

1 + 24 = 25

= 25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- 1 + 5}{2 ( - 2 )}, u_{2} = \frac{- 1 - 5}{2 ( - 2 )}

u = \frac{- 1 + 5}{2 ( - 2 )} : - 1

\frac{- 1 + 5}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 1 + 5}{- 2 \cdot 2}

数字相加/相减:

- 1 + 5 = 4

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

数字相乘:

2 \cdot 2 = 4

= \frac{4}{- 4}

使用分式法则:

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

= - \frac{4}{4}

使用法则

\frac{a}{a} = 1

= - 1

u = \frac{- 1 - 5}{2 ( - 2 )} : \frac{3}{2}

\frac{- 1 - 5}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 1 - 5}{- 2 \cdot 2}

数字相减:

- 1 - 5 = - 6

= \frac{- 6}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 6}{- 4}

使用分式法则:

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

= \frac{6}{4}

约分:

2

= \frac{3}{2}

二次方程组的解是:

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

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

验证解

找到无定义的点（奇点）:

u = 0

取

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

u = csc (x)

代回

csc (x) = - 1, csc (x) = \frac{3}{2}

csc (x) = - 1, csc (x) = \frac{3}{2}

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

csc (x) = - 1

csc (x) = - 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

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

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

csc (x) = \frac{3}{2}

使用反三角函数性质

csc (x) = \frac{3}{2}

csc (x) = \frac{3}{2}

的通解

csc (x) = a \Rightarrow x = arccsc (a) + 2 πn, x = π - arccsc (a) + 2 πn

x = arccsc (\frac{3}{2}) + 2 πn, x = π - arccsc (\frac{3}{2}) + 2 πn

x = arccsc (\frac{3}{2}) + 2 πn, x = π - arccsc (\frac{3}{2}) + 2 πn

合并所有解

x = \frac{3 π}{2} + 2 πn, x = arccsc (\frac{3}{2}) + 2 πn, x = π - arccsc (\frac{3}{2}) + 2 πn

以小数形式表示解

x = \frac{3 π}{2} + 2 πn, x = 0.72972 \dots + 2 πn, x = π - 0.72972 \dots + 2 πn

作图

流行的例子

cos (x) = a

3sin(x)+3cos(2x)=2

3 sin (x) + 3 cos (2 x) = 2

tan(3x+25)*160-1=0

tan (3 x + 2 5^{\circ}) \cdot 16 0^{\circ} - 1 = 0

5 cos (x) = 1

3cos(x)+5=7,0<= x<= 2pi

3 cos (x) + 5 = 7, 0 \leq x \leq 2 π