zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

2tan^2(θ)+5tan(θ)=0

解答

2 tan^{2} (θ) + 5 tan (θ) = 0

解答

θ = πn, θ = - 1.19028 \dots + πn

+1

度数

θ = 0^{\circ} + 18 0^{\circ} n, θ = - 68.19859 \dots^{\circ} + 18 0^{\circ} n

求解步骤

2 tan^{2} (θ) + 5 tan (θ) = 0

用替代法求解

2 tan^{2} (θ) + 5 tan (θ) = 0

令

: tan (θ) = u

2 u^{2} + 5 u = 0

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

2 u^{2} + 5 u = 0

使用求根公式求解

2 u^{2} + 5 u = 0

二次方程求根公式:

若

a = 2, b = 5, c = 0

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

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

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

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

将解分隔开

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

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

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

数字相加/相减:

- 5 + 5 = 0

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

数字相乘:

2 \cdot 2 = 4

= \frac{0}{4}

使用法则

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

= 0

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

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

数字相减:

- 5 - 5 = - 10

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 10}{4}

使用分式法则:

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

= - \frac{10}{4}

约分:

2

= - \frac{5}{2}

二次方程组的解是:

u = 0, u = - \frac{5}{2}

u = tan (θ)

代回

tan (θ) = 0, tan (θ) = - \frac{5}{2}

tan (θ) = 0, tan (θ) = - \frac{5}{2}

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{5}{2} : θ = arctan (- \frac{5}{2}) + πn

tan (θ) = - \frac{5}{2}

使用反三角函数性质

tan (θ) = - \frac{5}{2}

tan (θ) = - \frac{5}{2}

的通解

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

θ = arctan (- \frac{5}{2}) + πn

θ = arctan (- \frac{5}{2}) + πn

合并所有解

θ = πn, θ = arctan (- \frac{5}{2}) + πn

以小数形式表示解

θ = πn, θ = - 1.19028 \dots + πn

作图

流行的例子

cos(θ)-sqrt(3)sin(θ)=1

cos (θ) - 3 sin (θ) = 1

sec (x) = \frac{1}{2}

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

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

csch (x) = \frac{5}{12}

2sin^2(x)+sin(x)=1,0<= x<= 2pi

2 sin^{2} (x) + sin (x) = 1, 0 \leq x \leq 2 π