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受欢迎的三角函数 >

3/(tan(x))=(2tan(x))(2cos(x))

解答

\frac{3}{tan ( x )} = (2 tan (x)) (2 cos (x))

解答

x = 0.80515 \dots + 2 πn, x = 2 π - 0.80515 \dots + 2 πn

+1

度数

x = 46.13190 \dots^{\circ} + 36 0^{\circ} n, x = 313.86809 \dots^{\circ} + 36 0^{\circ} n

求解步骤

\frac{3}{tan ( x )} = (2 tan (x)) (2 cos (x))

两边减去

2 tan (x) 2 cos (x)

\frac{3}{tan ( x )} - 4 tan (x) cos (x) = 0

化简

\frac{3}{tan ( x )} - 4 tan (x) cos (x) : \frac{3 - 4 tan ^{2} ( x ) cos ( x )}{tan ( x )}

\frac{3}{tan ( x )} - 4 tan (x) cos (x)

将项转换为分式:

4 tan (x) cos (x) = \frac{4 tan ( x ) cos ( x ) tan ( x )}{tan ( x )}

= \frac{3}{tan ( x )} - \frac{4 tan ( x ) cos ( x ) tan ( x )}{tan ( x )}

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{3 - 4 tan ( x ) cos ( x ) tan ( x )}{tan ( x )}

3 - 4 tan (x) cos (x) tan (x) = 3 - 4 tan^{2} (x) cos (x)

3 - 4 tan (x) cos (x) tan (x)

4 tan (x) cos (x) tan (x) = 4 tan^{2} (x) cos (x)

4 tan (x) cos (x) tan (x)

使用指数法则:

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

tan (x) tan (x) = tan^{1 + 1} (x)

= 4 cos (x) tan^{1 + 1} (x)

数字相加:

1 + 1 = 2

= 4 cos (x) tan^{2} (x)

= 3 - 4 tan^{2} (x) cos (x)

= \frac{3 - 4 tan ^{2} ( x ) cos ( x )}{tan ( x )}

\frac{3 - 4 tan ^{2} ( x ) cos ( x )}{tan ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

3 - 4 tan^{2} (x) cos (x) = 0

用 sin, cos 表示

3 - 4 (\frac{sin ( x )}{cos ( x )})^{2} cos (x) = 0

化简

3 - 4 (\frac{sin ( x )}{cos ( x )})^{2} cos (x) : \frac{3 cos ( x ) - 4 sin ^{2} ( x )}{cos ( x )}

3 - 4 (\frac{sin ( x )}{cos ( x )})^{2} cos (x)

4 (\frac{sin ( x )}{cos ( x )})^{2} cos (x) = \frac{4 sin ^{2} ( x )}{cos ( x )}

4 (\frac{sin ( x )}{cos ( x )})^{2} cos (x)

(\frac{sin ( x )}{cos ( x )})^{2} = \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

(\frac{sin ( x )}{cos ( x )})^{2}

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

= 4 \cdot \frac{sin ^{2} ( x )}{cos ^{2} ( x )} cos (x)

分式相乘:

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

= \frac{sin ^{2} ( x ) \cdot 4 cos ( x )}{cos ^{2} ( x )}

约分:

cos (x)

= \frac{4 sin ^{2} ( x )}{cos ( x )}

= 3 - \frac{4 sin ^{2} ( x )}{cos ( x )}

将项转换为分式:

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

= \frac{3 cos ( x )}{cos ( x )} - \frac{4 sin ^{2} ( x )}{cos ( x )}

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{3 cos ( x ) - 4 sin ^{2} ( x )}{cos ( x )}

\frac{3 cos ( x ) - 4 sin ^{2} ( x )}{cos ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

3 cos (x) - 4 sin^{2} (x) = 0

两边加上

4 sin^{2} (x)

3 cos (x) = 4 sin^{2} (x)

两边进行平方

(3 cos (x))^{2} = (4 sin^{2} (x))^{2}

两边减去

(4 sin^{2} (x))^{2}

9 cos^{2} (x) - 16 sin^{4} (x) = 0

分解

9 cos^{2} (x) - 16 sin^{4} (x) : (3 cos (x) + 4 sin^{2} (x)) (3 cos (x) - 4 sin^{2} (x))

9 cos^{2} (x) - 16 sin^{4} (x)

将

9 cos^{2} (x) - 16 sin^{4} (x)

改写为

(3 cos (x))^{2} - (4 sin^{2} (x))^{2}

9 cos^{2} (x) - 16 sin^{4} (x)

将

9

改写为

3^{2}

= 3^{2} cos^{2} (x) - 16 sin^{4} (x)

将

16

改写为

4^{2}

= 3^{2} cos^{2} (x) - 4^{2} sin^{4} (x)

使用指数法则:

a^{b c} = (a^{b})^{c}

sin^{4} (x) = (sin^{2} (x))^{2}

= 3^{2} cos^{2} (x) - 4^{2} (sin^{2} (x))^{2}

使用指数法则:

a^{m} b^{m} = (ab)^{m}

3^{2} cos^{2} (x) = (3 cos (x))^{2}

= (3 cos (x))^{2} - 4^{2} (sin^{2} (x))^{2}

使用指数法则:

a^{m} b^{m} = (ab)^{m}

4^{2} (sin^{2} (x))^{2} = (4 sin^{2} (x))^{2}

= (3 cos (x))^{2} - (4 sin^{2} (x))^{2}

= (3 cos (x))^{2} - (4 sin^{2} (x))^{2}

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

(3 cos (x))^{2} - (4 sin^{2} (x))^{2} = (3 cos (x) + 4 sin^{2} (x)) (3 cos (x) - 4 sin^{2} (x))

= (3 cos (x) + 4 sin^{2} (x)) (3 cos (x) - 4 sin^{2} (x))

(3 cos (x) + 4 sin^{2} (x)) (3 cos (x) - 4 sin^{2} (x)) = 0

分别求解每个部分

3 cos (x) + 4 sin^{2} (x) = 0 or 3 cos (x) - 4 sin^{2} (x) = 0

3 cos (x) + 4 sin^{2} (x) = 0 : x = arccos (- \frac{- 3 + 73}{8}) + 2 πn, x = - arccos (- \frac{- 3 + 73}{8}) + 2 πn

3 cos (x) + 4 sin^{2} (x) = 0

使用三角恒等式改写

3 cos (x) + 4 sin^{2} (x)

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= 3 cos (x) + 4 (1 - cos^{2} (x))

(1 - cos^{2} (x)) \cdot 4 + 3 cos (x) = 0

用替代法求解

(1 - cos^{2} (x)) \cdot 4 + 3 cos (x) = 0

令

: cos (x) = u

(1 - u^{2}) \cdot 4 + 3 u = 0

(1 - u^{2}) \cdot 4 + 3 u = 0 : u = - \frac{- 3 + 73}{8}, u = \frac{3 + 73}{8}

(1 - u^{2}) \cdot 4 + 3 u = 0

展开

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

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

= 4 (1 - u^{2}) + 3 u

乘开

4 (1 - u^{2}) : 4 - 4 u^{2}

4 (1 - u^{2})

使用分配律:

a (b - c) = ab - a c

a = 4, b = 1, c = u^{2}

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

数字相乘:

4 \cdot 1 = 4

= 4 - 4 u^{2}

= 4 - 4 u^{2} + 3 u

4 - 4 u^{2} + 3 u = 0

改写成标准形式

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

- 4 u^{2} + 3 u + 4 = 0

使用求根公式求解

- 4 u^{2} + 3 u + 4 = 0

二次方程求根公式:

若

a = - 4, b = 3, c = 4

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

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

3^{2} - 4 (- 4) \cdot 4 = 73

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

使用法则

- (- a) = a

= 3^{2} + 4 \cdot 4 \cdot 4

数字相乘:

4 \cdot 4 \cdot 4 = 64

= 3^{2} + 64

3^{2} = 9

= 9 + 64

数字相加:

9 + 64 = 73

= 73

u_{1, 2} = \frac{- 3 \pm 73}{2 ( - 4 )}

将解分隔开

u_{1} = \frac{- 3 + 73}{2 ( - 4 )}, u_{2} = \frac{- 3 - 73}{2 ( - 4 )}

u = \frac{- 3 + 73}{2 ( - 4 )} : - \frac{- 3 + 73}{8}

\frac{- 3 + 73}{2 ( - 4 )}

去除括号:

(- a) = - a

= \frac{- 3 + 73}{- 2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 3 + 73}{- 8}

使用分式法则:

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

= - \frac{- 3 + 73}{8}

u = \frac{- 3 - 73}{2 ( - 4 )} : \frac{3 + 73}{8}

\frac{- 3 - 73}{2 ( - 4 )}

去除括号:

(- a) = - a

= \frac{- 3 - 73}{- 2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 3 - 73}{- 8}

使用分式法则:

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

- 3 - 73 = - (3 + 73)

= \frac{3 + 73}{8}

二次方程组的解是:

u = - \frac{- 3 + 73}{8}, u = \frac{3 + 73}{8}

u = cos (x)

代回

cos (x) = - \frac{- 3 + 73}{8}, cos (x) = \frac{3 + 73}{8}

cos (x) = - \frac{- 3 + 73}{8}, cos (x) = \frac{3 + 73}{8}

cos (x) = - \frac{- 3 + 73}{8} : x = arccos (- \frac{- 3 + 73}{8}) + 2 πn, x = - arccos (- \frac{- 3 + 73}{8}) + 2 πn

cos (x) = - \frac{- 3 + 73}{8}

使用反三角函数性质

cos (x) = - \frac{- 3 + 73}{8}

cos (x) = - \frac{- 3 + 73}{8}

的通解

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (- \frac{- 3 + 73}{8}) + 2 πn, x = - arccos (- \frac{- 3 + 73}{8}) + 2 πn

x = arccos (- \frac{- 3 + 73}{8}) + 2 πn, x = - arccos (- \frac{- 3 + 73}{8}) + 2 πn

cos (x) = \frac{3 + 73}{8} :

无解

cos (x) = \frac{3 + 73}{8}

- 1 \leq cos (x) \leq 1

无解

合并所有解

x = arccos (- \frac{- 3 + 73}{8}) + 2 πn, x = - arccos (- \frac{- 3 + 73}{8}) + 2 πn

3 cos (x) - 4 sin^{2} (x) = 0 : x = arccos (\frac{- 3 + 73}{8}) + 2 πn, x = 2 π - arccos (\frac{- 3 + 73}{8}) + 2 πn

3 cos (x) - 4 sin^{2} (x) = 0

使用三角恒等式改写

3 cos (x) - 4 sin^{2} (x)

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= 3 cos (x) - 4 (1 - cos^{2} (x))

- (1 - cos^{2} (x)) \cdot 4 + 3 cos (x) = 0

用替代法求解

- (1 - cos^{2} (x)) \cdot 4 + 3 cos (x) = 0

令

: cos (x) = u

- (1 - u^{2}) \cdot 4 + 3 u = 0

- (1 - u^{2}) \cdot 4 + 3 u = 0 : u = \frac{- 3 + 73}{8}, u = \frac{- 3 - 73}{8}

- (1 - u^{2}) \cdot 4 + 3 u = 0

展开

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

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

= - 4 (1 - u^{2}) + 3 u

乘开

- 4 (1 - u^{2}) : - 4 + 4 u^{2}

- 4 (1 - u^{2})

使用分配律:

a (b - c) = ab - a c

a = - 4, b = 1, c = u^{2}

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

使用加减运算法则

- (- a) = a

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

数字相乘:

4 \cdot 1 = 4

= - 4 + 4 u^{2}

= - 4 + 4 u^{2} + 3 u

- 4 + 4 u^{2} + 3 u = 0

改写成标准形式

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

4 u^{2} + 3 u - 4 = 0

使用求根公式求解

4 u^{2} + 3 u - 4 = 0

二次方程求根公式:

若

a = 4, b = 3, c = - 4

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

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

3^{2} - 4 \cdot 4 (- 4) = 73

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

使用法则

- (- a) = a

= 3^{2} + 4 \cdot 4 \cdot 4

数字相乘:

4 \cdot 4 \cdot 4 = 64

= 3^{2} + 64

3^{2} = 9

= 9 + 64

数字相加:

9 + 64 = 73

= 73

u_{1, 2} = \frac{- 3 \pm 73}{2 \cdot 4}

将解分隔开

u_{1} = \frac{- 3 + 73}{2 \cdot 4}, u_{2} = \frac{- 3 - 73}{2 \cdot 4}

u = \frac{- 3 + 73}{2 \cdot 4} : \frac{- 3 + 73}{8}

\frac{- 3 + 73}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 3 + 73}{8}

u = \frac{- 3 - 73}{2 \cdot 4} : \frac{- 3 - 73}{8}

\frac{- 3 - 73}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 3 - 73}{8}

二次方程组的解是:

u = \frac{- 3 + 73}{8}, u = \frac{- 3 - 73}{8}

u = cos (x)

代回

cos (x) = \frac{- 3 + 73}{8}, cos (x) = \frac{- 3 - 73}{8}

cos (x) = \frac{- 3 + 73}{8}, cos (x) = \frac{- 3 - 73}{8}

cos (x) = \frac{- 3 + 73}{8} : x = arccos (\frac{- 3 + 73}{8}) + 2 πn, x = 2 π - arccos (\frac{- 3 + 73}{8}) + 2 πn

cos (x) = \frac{- 3 + 73}{8}

使用反三角函数性质

cos (x) = \frac{- 3 + 73}{8}

cos (x) = \frac{- 3 + 73}{8}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{- 3 + 73}{8}) + 2 πn, x = 2 π - arccos (\frac{- 3 + 73}{8}) + 2 πn

x = arccos (\frac{- 3 + 73}{8}) + 2 πn, x = 2 π - arccos (\frac{- 3 + 73}{8}) + 2 πn

cos (x) = \frac{- 3 - 73}{8} :

无解

cos (x) = \frac{- 3 - 73}{8}

- 1 \leq cos (x) \leq 1

无解

合并所有解

x = arccos (\frac{- 3 + 73}{8}) + 2 πn, x = 2 π - arccos (\frac{- 3 + 73}{8}) + 2 πn

合并所有解

x = arccos (- \frac{- 3 + 73}{8}) + 2 πn, x = - arccos (- \frac{- 3 + 73}{8}) + 2 πn, x = arccos (\frac{- 3 + 73}{8}) + 2 πn, x = 2 π - arccos (\frac{- 3 + 73}{8}) + 2 πn

将解代入原方程进行验证

将它们代入

\frac{3}{tan ( x )} = 2 tan (x) 2 cos (x)

检验解是否符合

去除与方程不符的解。

检验

arccos (- \frac{- 3 + 73}{8}) + 2 πn

的解:

假

arccos (- \frac{- 3 + 73}{8}) + 2 πn

代入

n = 1

arccos (- \frac{- 3 + 73}{8}) + 2 π 1

对于

\frac{3}{tan ( x )} = 2 tan (x) 2 cos (x)

代入

x = arccos (- \frac{- 3 + 73}{8}) + 2 π 1

\frac{3}{tan ( arccos ( - \frac{- 3 + 73}{8} ) + 2 π 1 )} = 2 tan (arccos (- \frac{- 3 + 73}{8}) + 2 π 1) \cdot 2 cos (arccos (- \frac{- 3 + 73}{8}) + 2 π 1)

整理后得

- 2.88374 \dots = 2.88374 \dots

\Rightarrow 假

检验

- arccos (- \frac{- 3 + 73}{8}) + 2 πn

的解:

假

- arccos (- \frac{- 3 + 73}{8}) + 2 πn

代入

n = 1

- arccos (- \frac{- 3 + 73}{8}) + 2 π 1

对于

\frac{3}{tan ( x )} = 2 tan (x) 2 cos (x)

代入

x = - arccos (- \frac{- 3 + 73}{8}) + 2 π 1

\frac{3}{tan ( - arccos ( - \frac{- 3 + 73}{8} ) + 2 π 1 )} = 2 tan (- arccos (- \frac{- 3 + 73}{8}) + 2 π 1) \cdot 2 cos (- arccos (- \frac{- 3 + 73}{8}) + 2 π 1)

整理后得

2.88374 \dots = - 2.88374 \dots

\Rightarrow 假

检验

arccos (\frac{- 3 + 73}{8}) + 2 πn

的解:

真

arccos (\frac{- 3 + 73}{8}) + 2 πn

代入

n = 1

arccos (\frac{- 3 + 73}{8}) + 2 π 1

对于

\frac{3}{tan ( x )} = 2 tan (x) 2 cos (x)

代入

x = arccos (\frac{- 3 + 73}{8}) + 2 π 1

\frac{3}{tan ( arccos ( \frac{- 3 + 73}{8} ) + 2 π 1 )} = 2 tan (arccos (\frac{- 3 + 73}{8}) + 2 π 1) \cdot 2 cos (arccos (\frac{- 3 + 73}{8}) + 2 π 1)

整理后得

2.88374 \dots = 2.88374 \dots

\Rightarrow 真

检验

2 π - arccos (\frac{- 3 + 73}{8}) + 2 πn

的解:

真

2 π - arccos (\frac{- 3 + 73}{8}) + 2 πn

代入

n = 1

2 π - arccos (\frac{- 3 + 73}{8}) + 2 π 1

对于

\frac{3}{tan ( x )} = 2 tan (x) 2 cos (x)

代入

x = 2 π - arccos (\frac{- 3 + 73}{8}) + 2 π 1

\frac{3}{tan ( 2 π - arccos ( \frac{- 3 + 73}{8} ) + 2 π 1 )} = 2 tan (2 π - arccos (\frac{- 3 + 73}{8}) + 2 π 1) \cdot 2 cos (2 π - arccos (\frac{- 3 + 73}{8}) + 2 π 1)

整理后得

- 2.88374 \dots = - 2.88374 \dots

\Rightarrow 真

x = arccos (\frac{- 3 + 73}{8}) + 2 πn, x = 2 π - arccos (\frac{- 3 + 73}{8}) + 2 πn

以小数形式表示解

x = 0.80515 \dots + 2 πn, x = 2 π - 0.80515 \dots + 2 πn

作图

流行的例子

cos(θ)=-4/5 ,sin(θ)

cos (θ) = - \frac{4}{5}, sin (θ)

5sin(4x)=4cos(2x)

5 sin (4 x) = 4 cos (2 x)

arccosh (x) = 1

solvefor θ,sin(3θ)= 1/2

so l v e f or θ, sin (3 θ) = \frac{1}{2}

0 = - 0.5 sin (\frac{x}{5})