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受欢迎的三角函数 >

tan^2(x)+5tan(x)=0

解答

tan^{2} (x) + 5 tan (x) = 0

解答

x = πn, x = - 1.37340 \dots + πn

+1

度数

x = 0^{\circ} + 18 0^{\circ} n, x = - 78.69006 \dots^{\circ} + 18 0^{\circ} n

求解步骤

tan^{2} (x) + 5 tan (x) = 0

用替代法求解

tan^{2} (x) + 5 tan (x) = 0

令

: tan (x) = u

u^{2} + 5 u = 0

u^{2} + 5 u = 0 : u = 0, u = - 5

u^{2} + 5 u = 0

使用求根公式求解

u^{2} + 5 u = 0

二次方程求根公式:

若

a = 1, b = 5, c = 0

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

5^{2} - 4 \cdot 1 \cdot 0 = 5

5^{2} - 4 \cdot 1 \cdot 0

使用法则

0 \cdot a = 0

= 5^{2} - 0

5^{2} - 0 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 5

u_{1, 2} = \frac{- 5 \pm 5}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- 5 + 5}{2 \cdot 1}, u_{2} = \frac{- 5 - 5}{2 \cdot 1}

u = \frac{- 5 + 5}{2 \cdot 1} : 0

\frac{- 5 + 5}{2 \cdot 1}

数字相加/相减:

- 5 + 5 = 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

u = \frac{- 5 - 5}{2 \cdot 1} : - 5

\frac{- 5 - 5}{2 \cdot 1}

数字相减:

- 5 - 5 = - 10

= \frac{- 10}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 10}{2}

使用分式法则:

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

= - \frac{10}{2}

数字相除:

\frac{10}{2} = 5

= - 5

二次方程组的解是:

u = 0, u = - 5

u = tan (x)

代回

tan (x) = 0, tan (x) = - 5

tan (x) = 0, tan (x) = - 5

tan (x) = 0 : x = πn

tan (x) = 0

tan (x) = 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}

x = 0 + πn

x = 0 + πn

解

x = 0 + πn : x = πn

x = 0 + πn

0 + πn = πn

x = πn

x = πn

tan (x) = - 5 : x = arctan (- 5) + πn

tan (x) = - 5

使用反三角函数性质

tan (x) = - 5

tan (x) = - 5

的通解

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

x = arctan (- 5) + πn

x = arctan (- 5) + πn

合并所有解

x = πn, x = arctan (- 5) + πn

以小数形式表示解

x = πn, x = - 1.37340 \dots + πn

作图

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