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受欢迎的三角函数 >

7+csc(2x)=3cot^2(2x)

解答

7 + csc (2 x) = 3 cot^{2} (2 x)

解答

x = - \frac{0.64350 \dots}{2} + πn, x = \frac{π}{2} + \frac{0.64350 \dots}{2} + πn, x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

+1

度数

x = - 18.43494 \dots^{\circ} + 18 0^{\circ} n, x = 108.43494 \dots^{\circ} + 18 0^{\circ} n, x = 1 5^{\circ} + 18 0^{\circ} n, x = 7 5^{\circ} + 18 0^{\circ} n

求解步骤

7 + csc (2 x) = 3 cot^{2} (2 x)

两边减去

3 cot^{2} (2 x)

7 + csc (2 x) - 3 cot^{2} (2 x) = 0

使用三角恒等式改写

7 + csc (2 x) - 3 cot^{2} (2 x)

使用毕达哥拉斯恒等式:

1 + cot^{2} (x) = csc^{2} (x)

cot^{2} (x) = csc^{2} (x) - 1

= 7 + csc (2 x) - 3 (csc^{2} (2 x) - 1)

化简

7 + csc (2 x) - 3 (csc^{2} (2 x) - 1) : csc (2 x) - 3 csc^{2} (2 x) + 10

7 + csc (2 x) - 3 (csc^{2} (2 x) - 1)

乘开

- 3 (csc^{2} (2 x) - 1) : - 3 csc^{2} (2 x) + 3

- 3 (csc^{2} (2 x) - 1)

使用分配律:

a (b - c) = ab - a c

a = - 3, b = csc^{2} (2 x), c = 1

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

使用加减运算法则

- (- a) = a

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

数字相乘:

3 \cdot 1 = 3

= - 3 csc^{2} (2 x) + 3

= 7 + csc (2 x) - 3 csc^{2} (2 x) + 3

化简

7 + csc (2 x) - 3 csc^{2} (2 x) + 3 : csc (2 x) - 3 csc^{2} (2 x) + 10

7 + csc (2 x) - 3 csc^{2} (2 x) + 3

对同类项分组

= csc (2 x) - 3 csc^{2} (2 x) + 7 + 3

数字相加:

7 + 3 = 10

= csc (2 x) - 3 csc^{2} (2 x) + 10

= csc (2 x) - 3 csc^{2} (2 x) + 10

= csc (2 x) - 3 csc^{2} (2 x) + 10

10 + csc (2 x) - 3 csc^{2} (2 x) = 0

用替代法求解

10 + csc (2 x) - 3 csc^{2} (2 x) = 0

令

: csc (2 x) = u

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

10 + u - 3 u^{2} = 0 : u = - \frac{5}{3}, u = 2

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

改写成标准形式

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

- 3 u^{2} + u + 10 = 0

使用求根公式求解

- 3 u^{2} + u + 10 = 0

二次方程求根公式:

若

a = - 3, b = 1, c = 10

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 3 ) \cdot 10}{2 ( - 3 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 3 ) \cdot 10}{2 ( - 3 )}

1^{2} - 4 (- 3) \cdot 10 = 11

1^{2} - 4 (- 3) \cdot 10

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 3) \cdot 10

使用法则

- (- a) = a

= 1 + 4 \cdot 3 \cdot 10

数字相乘:

4 \cdot 3 \cdot 10 = 120

= 1 + 120

数字相加:

1 + 120 = 121

= 121

因式分解数字:

121 = 1 1^{2}

= 1 1^{2}

使用根式运算法则:

n a^{n} = a

1 1^{2} = 11

= 11

u_{1, 2} = \frac{- 1 \pm 11}{2 ( - 3 )}

将解分隔开

u_{1} = \frac{- 1 + 11}{2 ( - 3 )}, u_{2} = \frac{- 1 - 11}{2 ( - 3 )}

u = \frac{- 1 + 11}{2 ( - 3 )} : - \frac{5}{3}

\frac{- 1 + 11}{2 ( - 3 )}

去除括号:

(- a) = - a

= \frac{- 1 + 11}{- 2 \cdot 3}

数字相加/相减:

- 1 + 11 = 10

= \frac{10}{- 2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{10}{- 6}

使用分式法则:

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

= - \frac{10}{6}

约分:

2

= - \frac{5}{3}

u = \frac{- 1 - 11}{2 ( - 3 )} : 2

\frac{- 1 - 11}{2 ( - 3 )}

去除括号:

(- a) = - a

= \frac{- 1 - 11}{- 2 \cdot 3}

数字相减:

- 1 - 11 = - 12

= \frac{- 12}{- 2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{- 12}{- 6}

使用分式法则:

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

= \frac{12}{6}

数字相除:

\frac{12}{6} = 2

= 2

二次方程组的解是:

u = - \frac{5}{3}, u = 2

u = csc (2 x)

代回

csc (2 x) = - \frac{5}{3}, csc (2 x) = 2

csc (2 x) = - \frac{5}{3}, csc (2 x) = 2

csc (2 x) = - \frac{5}{3} : x = - \frac{arccsc ( \frac{5}{3} )}{2} + πn, x = \frac{π}{2} + \frac{arccsc ( \frac{5}{3} )}{2} + πn

csc (2 x) = - \frac{5}{3}

使用反三角函数性质

csc (2 x) = - \frac{5}{3}

csc (2 x) = - \frac{5}{3}

的通解

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

2 x = arccsc (- \frac{5}{3}) + 2 πn, 2 x = π + arccsc (\frac{5}{3}) + 2 πn

2 x = arccsc (- \frac{5}{3}) + 2 πn, 2 x = π + arccsc (\frac{5}{3}) + 2 πn

解

2 x = arccsc (- \frac{5}{3}) + 2 πn : x = - \frac{arccsc ( \frac{5}{3} )}{2} + πn

2 x = arccsc (- \frac{5}{3}) + 2 πn

化简

arccsc (- \frac{5}{3}) + 2 πn : - arccsc (\frac{5}{3}) + 2 πn

arccsc (- \frac{5}{3}) + 2 πn

利用以下特性:

arccsc (- x) = - arccsc (x)

arccsc (- \frac{5}{3}) = - arccsc (\frac{5}{3})

= - arccsc (\frac{5}{3}) + 2 πn

2 x = - arccsc (\frac{5}{3}) + 2 πn

两边除以

2

2 x = - arccsc (\frac{5}{3}) + 2 πn

两边除以

2

\frac{2 x}{2} = - \frac{arccsc ( \frac{5}{3} )}{2} + \frac{2 πn}{2}

化简

x = - \frac{arccsc ( \frac{5}{3} )}{2} + πn

x = - \frac{arccsc ( \frac{5}{3} )}{2} + πn

解

2 x = π + arccsc (\frac{5}{3}) + 2 πn : x = \frac{π}{2} + \frac{arccsc ( \frac{5}{3} )}{2} + πn

2 x = π + arccsc (\frac{5}{3}) + 2 πn

两边除以

2

2 x = π + arccsc (\frac{5}{3}) + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{π}{2} + \frac{arccsc ( \frac{5}{3} )}{2} + \frac{2 πn}{2}

化简

x = \frac{π}{2} + \frac{arccsc ( \frac{5}{3} )}{2} + πn

x = \frac{π}{2} + \frac{arccsc ( \frac{5}{3} )}{2} + πn

x = - \frac{arccsc ( \frac{5}{3} )}{2} + πn, x = \frac{π}{2} + \frac{arccsc ( \frac{5}{3} )}{2} + πn

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{arccsc ( \frac{5}{3} )}{2} + πn, x = \frac{π}{2} + \frac{arccsc ( \frac{5}{3} )}{2} + πn, x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

以小数形式表示解

x = - \frac{0.64350 \dots}{2} + πn, x = \frac{π}{2} + \frac{0.64350 \dots}{2} + πn, x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

作图

流行的例子

cos(θ)=cos(pi/(13))

cos (θ) = cos (\frac{π}{13})

sin(2x+pi/6)=-1

sin (2 x + \frac{π}{6}) = - 1

tan (x) = 2 π

2sin^2(x)-sqrt(2)=0

2 sin^{2} (x) - 2 = 0

solvefor x,z=sin(5x)y^3

so l v e f or x, z = sin (5 x) y^{3}