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受欢迎的三角函数 >

3tan^3(θ)=tan(θ)

解答

3 tan^{3} (θ) = tan (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^{3} (θ) = tan (θ)

用替代法求解

3 tan^{3} (θ) = tan (θ)

令

: tan (θ) = u

3 u^{3} = u

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

3 u^{3} = u

将

u

para o lado esquerdo

3 u^{3} = u

两边减去

u

3 u^{3} - u = u - u

化简

3 u^{3} - u = 0

3 u^{3} - u = 0

因式分解

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

3 u^{3} - u

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 = 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

csc^{2} (x) - 4 = 0

cos (\frac{π}{12})

sin^{2} (x) - 1 = 0

cot (6 0^{\circ})

sec^{2} (\frac{π}{2})