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受欢迎的三角函数 >

tan^2(θ)-2tan(θ)=0

解答

tan^{2} (θ) - 2 tan (θ) = 0

解答

θ = 1.10714 \dots + πn, θ = πn

+1

度数

θ = 63.43494 \dots^{\circ} + 18 0^{\circ} n, θ = 0^{\circ} + 18 0^{\circ} n

求解步骤

tan^{2} (θ) - 2 tan (θ) = 0

用替代法求解

tan^{2} (θ) - 2 tan (θ) = 0

令

: tan (θ) = u

u^{2} - 2 u = 0

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

u^{2} - 2 u = 0

使用求根公式求解

u^{2} - 2 u = 0

二次方程求根公式:

若

a = 1, b = - 2, c = 0

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

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

(- 2)^{2} - 4 \cdot 1 \cdot 0

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 2)^{2} = 2^{2}

= 2^{2} - 4 \cdot 1 \cdot 0

使用法则

0 \cdot a = 0

= 2^{2} - 0

2^{2} - 0 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 2

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

将解分隔开

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

u = \frac{- ( - 2 ) + 2}{2 \cdot 1} : 2

\frac{- ( - 2 ) + 2}{2 \cdot 1}

使用法则

- (- a) = a

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

数字相加:

2 + 2 = 4

= \frac{4}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{4}{2}

数字相除:

\frac{4}{2} = 2

= 2

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

\frac{- ( - 2 ) - 2}{2 \cdot 1}

使用法则

- (- a) = a

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

数字相减:

2 - 2 = 0

= \frac{0}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{0}{2}

使用法则

\frac{0}{a} = 0, a \neq = 0

= 0

二次方程组的解是:

u = 2, u = 0

u = tan (θ)

代回

tan (θ) = 2, tan (θ) = 0

tan (θ) = 2, tan (θ) = 0

tan (θ) = 2 : θ = arctan (2) + πn

tan (θ) = 2

使用反三角函数性质

tan (θ) = 2

tan (θ) = 2

的通解

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

θ = arctan (2) + πn

θ = arctan (2) + πn

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

合并所有解

θ = arctan (2) + πn, θ = πn

以小数形式表示解

θ = 1.10714 \dots + πn, θ = πn

作图

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