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受欢迎的三角函数 >

cot(x)*tan(2x)=3

解答

cot (x) \cdot tan (2 x) = 3

解答

x = 0.52359 \dots + πn, x = - 0.52359 \dots + πn

+1

度数

x = 3 0^{\circ} + 18 0^{\circ} n, x = - 3 0^{\circ} + 18 0^{\circ} n

求解步骤

cot (x) tan (2 x) = 3

两边减去

3

cot (x) tan (2 x) - 3 = 0

使用三角恒等式改写

- 3 + cot (x) tan (2 x)

使用基本三角恒等式:

cot (x) = \frac{1}{tan ( x )}

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

\frac{1}{tan ( x )} tan (2 x) = \frac{tan ( 2 x )}{tan ( x )}

\frac{1}{tan ( x )} tan (2 x)

分式相乘:

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

= \frac{1 \cdot tan ( 2 x )}{tan ( x )}

乘以:

1 \cdot tan (2 x) = tan (2 x)

= \frac{tan ( 2 x )}{tan ( x )}

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

使用倍角公式:

tan (2 x) = \frac{2 tan ( x )}{1 - tan ^{2} ( x )}

= - 3 + \frac{\frac{2 t a n ( x )}{1 - t a n ^{2} ( x )}}{tan ( x )}

\frac{\frac{2 t a n ( x )}{1 - t a n ^{2} ( x )}}{tan ( x )} = \frac{2}{1 - tan ^{2} ( x )}

\frac{\frac{2 t a n ( x )}{1 - t a n ^{2} ( x )}}{tan ( x )}

使用分式法则:

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

= \frac{2 tan ( x )}{( 1 - tan ^{2} ( x ) ) tan ( x )}

约分:

tan (x)

= \frac{2}{1 - tan ^{2} ( x )}

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

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

用替代法求解

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

令

: tan (x) = u

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

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

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

在两边乘以

1 - u^{2}

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

在两边乘以

1 - u^{2}

- 3 (1 - u^{2}) + \frac{2}{1 - u ^{2}} (1 - u^{2}) = 0 \cdot (1 - u^{2})

化简

- 3 (1 - u^{2}) + \frac{2}{1 - u ^{2}} (1 - u^{2}) = 0 \cdot (1 - u^{2})

化简

\frac{2}{1 - u ^{2}} (1 - u^{2}) : 2

\frac{2}{1 - u ^{2}} (1 - u^{2})

分式相乘:

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

= \frac{2 ( 1 - u ^{2} )}{1 - u ^{2}}

约分:

1 - u^{2}

= 2

化简

0 \cdot (1 - u^{2}) : 0

0 \cdot (1 - u^{2})

使用法则

0 \cdot a = 0

= 0

- 3 (1 - u^{2}) + 2 = 0

- 3 (1 - u^{2}) + 2 = 0

- 3 (1 - u^{2}) + 2 = 0

解

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

- 3 (1 - u^{2}) + 2 = 0

将

2

到右边

- 3 (1 - u^{2}) + 2 = 0

两边减去

2

- 3 (1 - u^{2}) + 2 - 2 = 0 - 2

化简

- 3 (1 - u^{2}) = - 2

- 3 (1 - u^{2}) = - 2

两边除以

- 3

- 3 (1 - u^{2}) = - 2

两边除以

- 3

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

化简

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

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

将

1

到右边

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

两边减去

1

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

化简

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

化简

1 - u^{2} - 1 : - u^{2}

1 - u^{2} - 1

同类项相加:

1 - 1 = 0

= - u^{2}

化简

\frac{2}{3} - 1 : - \frac{1}{3}

\frac{2}{3} - 1

将项转换为分式:

1 = \frac{1 \cdot 3}{3}

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

因为分母相等，所以合并分式:

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

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

- 1 \cdot 3 + 2 = - 1

- 1 \cdot 3 + 2

数字相乘:

1 \cdot 3 = 3

= - 3 + 2

数字相加/相减:

- 3 + 2 = - 1

= - 1

= \frac{- 1}{3}

使用分式法则:

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

= - \frac{1}{3}

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

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

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

两边除以

- 1

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

两边除以

- 1

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

化简

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

化简

\frac{- u ^{2}}{- 1} : u^{2}

\frac{- u ^{2}}{- 1}

使用分式法则:

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

= \frac{u ^{2}}{1}

使用法则

\frac{a}{1} = a

= u^{2}

化简

\frac{- \frac{1}{3}}{- 1} : \frac{1}{3}

\frac{- \frac{1}{3}}{- 1}

使用分式法则:

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

= \frac{\frac{1}{3}}{1}

使用分式法则:

\frac{a}{1} = a

= \frac{1}{3}

u^{2} = \frac{1}{3}

u^{2} = \frac{1}{3}

u^{2} = \frac{1}{3}

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

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

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

验证解

找到无定义的点（奇点）:

u = 1, u = - 1

取

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

的分母，令其等于零

解

1 - u^{2} = 0 : u = 1, u = - 1

1 - u^{2} = 0

将

1

到右边

1 - u^{2} = 0

两边减去

1

1 - u^{2} - 1 = 0 - 1

化简

- u^{2} = - 1

- u^{2} = - 1

两边除以

- 1

- u^{2} = - 1

两边除以

- 1

\frac{- u ^{2}}{- 1} = \frac{- 1}{- 1}

化简

u^{2} = 1

u^{2} = 1

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

使用根式运算法则:

1 = 1

= 1

- 1 = - 1

- 1

使用根式运算法则:

1 = 1

1 = 1

= - 1

u = 1, u = - 1

以下点无定义

u = 1, u = - 1

将不在定义域的点与解相综合:

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

u = tan (x)

代回

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

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

tan (x) = \frac{1}{3} : x = arctan (\frac{1}{3}) + πn

tan (x) = \frac{1}{3}

使用反三角函数性质

tan (x) = \frac{1}{3}

tan (x) = \frac{1}{3}

的通解

tan (x) = a \Rightarrow x = arctan (a) + πn

x = arctan (\frac{1}{3}) + πn

x = arctan (\frac{1}{3}) + πn

tan (x) = - \frac{1}{3} : x = arctan (- \frac{1}{3}) + πn

tan (x) = - \frac{1}{3}

使用反三角函数性质

tan (x) = - \frac{1}{3}

tan (x) = - \frac{1}{3}

的通解

tan (x) = - a \Rightarrow x = arctan (- a) + πn

x = arctan (- \frac{1}{3}) + πn

x = arctan (- \frac{1}{3}) + πn

合并所有解

x = arctan (\frac{1}{3}) + πn, x = arctan (- \frac{1}{3}) + πn

以小数形式表示解

x = 0.52359 \dots + πn, x = - 0.52359 \dots + πn

作图

流行的例子

1*sin(35)=1.33*sin(θ)

1 \cdot sin (3 5^{\circ}) = 1.33 \cdot sin (θ)

cos (3 A) = 1

cos(θ)=(-0.68159)

cos (θ) = (- 0.68159)

cot(x+(5pi)/6)=sqrt(3)

cot (x + \frac{5 π}{6}) = 3

solvefor θ,y=sin(θ)

so l v e f or θ, y = sin (θ)