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受欢迎的三角函数 >

3tan(θ)=tan(2θ)

解答

3 tan (θ) = tan (2 θ)

解答

θ = πn, θ = \frac{5 π}{6} + πn, θ = \frac{π}{6} + πn

+1

度数

θ = 0^{\circ} + 18 0^{\circ} n, θ = 15 0^{\circ} + 18 0^{\circ} n, θ = 3 0^{\circ} + 18 0^{\circ} n

求解步骤

3 tan (θ) = tan (2 θ)

两边减去

tan (2 θ)

3 tan (θ) - tan (2 θ) = 0

使用三角恒等式改写

- tan (2 θ) + 3 tan (θ)

使用倍角公式:

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

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

化简

- \frac{2 tan ( θ )}{1 - tan ^{2} ( θ )} + 3 tan (θ) : \frac{tan ( θ ) - 3 tan ^{3} ( θ )}{1 - tan ^{2} ( θ )}

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

将项转换为分式:

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

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

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{- 2 tan ( θ ) + 3 tan ( θ ) ( 1 - tan ^{2} ( θ ) )}{1 - tan ^{2} ( θ )}

乘开

- 2 tan (θ) + 3 tan (θ) (1 - tan^{2} (θ)) : tan (θ) - 3 tan^{3} (θ)

- 2 tan (θ) + 3 tan (θ) (1 - tan^{2} (θ))

乘开

3 tan (θ) (1 - tan^{2} (θ)) : 3 tan (θ) - 3 tan^{3} (θ)

3 tan (θ) (1 - tan^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = 3 tan (θ), b = 1, c = tan^{2} (θ)

= 3 tan (θ) \cdot 1 - 3 tan (θ) tan^{2} (θ)

= 3 \cdot 1 \cdot tan (θ) - 3 tan^{2} (θ) tan (θ)

化简

3 \cdot 1 \cdot tan (θ) - 3 tan^{2} (θ) tan (θ) : 3 tan (θ) - 3 tan^{3} (θ)

3 \cdot 1 \cdot tan (θ) - 3 tan^{2} (θ) tan (θ)

3 \cdot 1 \cdot tan (θ) = 3 tan (θ)

3 \cdot 1 \cdot tan (θ)

数字相乘:

3 \cdot 1 = 3

= 3 tan (θ)

3 tan^{2} (θ) tan (θ) = 3 tan^{3} (θ)

3 tan^{2} (θ) tan (θ)

使用指数法则:

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

tan^{2} (θ) tan (θ) = tan^{2 + 1} (θ)

= 3 tan^{2 + 1} (θ)

数字相加:

2 + 1 = 3

= 3 tan^{3} (θ)

= 3 tan (θ) - 3 tan^{3} (θ)

= 3 tan (θ) - 3 tan^{3} (θ)

= - 2 tan (θ) + 3 tan (θ) - 3 tan^{3} (θ)

同类项相加:

- 2 tan (θ) + 3 tan (θ) = tan (θ)

= tan (θ) - 3 tan^{3} (θ)

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

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

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

用替代法求解

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

令

: tan (θ) = u

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

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

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

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

u - 3 u^{3} = 0

解

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

u - 3 u^{3} = 0

因式分解

u - 3 u^{3} : - u (3 u + 1) (3 u - 1)

u - 3 u^{3}

因式分解出通项

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

- 3 u^{3} + u

使用指数法则:

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

u^{3} = u^{2} u

= - 3 u^{2} u + u

因式分解出通项

- u

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

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

分解

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

3 u^{2} - 1

将

3 u^{2} - 1

改写为

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

3 u^{2} - 1

使用根式运算法则:

a = (a)^{2}

3 = (3)^{2}

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

将

1

改写为

1^{2}

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

使用指数法则:

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

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

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

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

使用平方差公式:

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

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

= (3 u + 1) (3 u - 1)

= - u (3 u + 1) (3 u - 1)

- u (3 u + 1) (3 u - 1) = 0

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

u = 0 or 3 u + 1 = 0 or 3 u - 1 = 0

解

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

3 u + 1 = 0

将

1

到右边

3 u + 1 = 0

两边减去

1

3 u + 1 - 1 = 0 - 1

化简

3 u = - 1

3 u = - 1

两边除以

3

3 u = - 1

两边除以

3

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

化简

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

化简

\frac{3 u}{3} : u

\frac{3 u}{3}

约分:

3

= u

化简

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

\frac{- 1}{3}

使用分式法则:

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

= - \frac{1}{3}

- \frac{1}{3}

有理化:

- \frac{3}{3}

- \frac{1}{3}

乘以共轭根式

\frac{3}{3}

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

1 \cdot 3 = 3

33 = 3

33

使用根式运算法则:

a a = a

33 = 3

= 3

= - \frac{3}{3}

= - \frac{3}{3}

u = - \frac{3}{3}

u = - \frac{3}{3}

u = - \frac{3}{3}

解

3 u - 1 = 0 : u = \frac{3}{3}

3 u - 1 = 0

将

1

到右边

3 u - 1 = 0

两边加上

1

3 u - 1 + 1 = 0 + 1

化简

3 u = 1

3 u = 1

两边除以

3

3 u = 1

两边除以

3

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

化简

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

化简

\frac{3 u}{3} : u

\frac{3 u}{3}

约分:

3

= u

化简

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

\frac{1}{3}

乘以共轭根式

\frac{3}{3}

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

1 \cdot 3 = 3

33 = 3

33

使用根式运算法则:

a a = a

33 = 3

= 3

= \frac{3}{3}

u = \frac{3}{3}

u = \frac{3}{3}

u = \frac{3}{3}

解为

u = 0, u = - \frac{3}{3}, u = \frac{3}{3}

u = 0, u = - \frac{3}{3}, u = \frac{3}{3}

验证解

找到无定义的点（奇点）:

u = 1, u = - 1

取

\frac{u - 3 u ^{3}}{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 = 0, u = - \frac{3}{3}, u = \frac{3}{3}

u = tan (θ)

代回

tan (θ) = 0, tan (θ) = - \frac{3}{3}, tan (θ) = \frac{3}{3}

tan (θ) = 0, tan (θ) = - \frac{3}{3}, tan (θ) = \frac{3}{3}

tan (θ) = 0 : θ = πn

tan (θ) = 0

tan (θ) = 0

的通解

tan (x)

周期表（周期为

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = 0 + πn

θ = 0 + πn

解

θ = 0 + πn : θ = πn

θ = 0 + πn

0 + πn = πn

θ = πn

θ = πn

tan (θ) = - \frac{3}{3} : θ = \frac{5 π}{6} + πn

tan (θ) = - \frac{3}{3}

tan (θ) = - \frac{3}{3}

的通解

tan (x)

周期表（周期为

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = \frac{5 π}{6} + πn

θ = \frac{5 π}{6} + πn

tan (θ) = \frac{3}{3} : θ = \frac{π}{6} + πn

tan (θ) = \frac{3}{3}

tan (θ) = \frac{3}{3}

的通解

tan (x)

周期表（周期为

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = \frac{π}{6} + πn

θ = \frac{π}{6} + πn

合并所有解

θ = πn, θ = \frac{5 π}{6} + πn, θ = \frac{π}{6} + πn

作图

流行的例子

5 cos (x) - 4 = 0

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

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

sin (x) = 14

2sec^2(x)-5sec(x)+2=0,0<= x<= 360

2 sec^{2} (x) - 5 sec (x) + 2 = 0, 0^{\circ} \leq x \leq 36 0^{\circ}

2sin(5x)+sqrt(3)=0,0<= x<= 2pi

2 sin (5 x) + 3 = 0, 0 \leq x \leq 2 π