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 >

3tan^2(x)-sqrt(3)tan(x)=0

解答

3 tan^{2} (x) - 3 tan (x) = 0

解答

x = \frac{π}{6} + πn, x = πn

+1

度数

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

求解步骤

3 tan^{2} (x) - 3 tan (x) = 0

用替代法求解

3 tan^{2} (x) - 3 tan (x) = 0

令

: tan (x) = u

3 u^{2} - 3 u = 0

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

3 u^{2} - 3 u = 0

使用求根公式求解

3 u^{2} - 3 u = 0

二次方程求根公式:

若

a = 3, b = - 3, c = 0

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 3 \cdot 0}{2 \cdot 3}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 3 \cdot 0}{2 \cdot 3}

(- 3)^{2} - 4 \cdot 3 \cdot 0 = 3

(- 3)^{2} - 4 \cdot 3 \cdot 0

(- 3)^{2} = 3

(- 3)^{2}

使用指数法则:

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

若

n

是偶数

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

= (3)^{2}

使用根式运算法则:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= 3^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

约分:

2

= 1

= 3

4 \cdot 3 \cdot 0 = 0

4 \cdot 3 \cdot 0

使用法则

0 \cdot a = 0

= 0

= 3 - 0

数字相减:

3 - 0 = 3

= 3

u_{1, 2} = \frac{- ( - 3 ) \pm 3}{2 \cdot 3}

将解分隔开

u_{1} = \frac{- ( - 3 ) + 3}{2 \cdot 3}, u_{2} = \frac{- ( - 3 ) - 3}{2 \cdot 3}

u = \frac{- ( - 3 ) + 3}{2 \cdot 3} : \frac{3}{3}

\frac{- ( - 3 ) + 3}{2 \cdot 3}

使用法则

- (- a) = a

= \frac{3 + 3}{2 \cdot 3}

同类项相加:

3 + 3 = 23

= \frac{2 3}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{2 3}{6}

约分:

2

= \frac{3}{3}

u = \frac{- ( - 3 ) - 3}{2 \cdot 3} : 0

\frac{- ( - 3 ) - 3}{2 \cdot 3}

使用法则

- (- a) = a

= \frac{3 - 3}{2 \cdot 3}

同类项相加:

3 - 3 = 0

= \frac{0}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{0}{6}

使用法则

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

= 0

二次方程组的解是:

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

u = tan (x)

代回

tan (x) = \frac{3}{3}, tan (x) = 0

tan (x) = \frac{3}{3}, tan (x) = 0

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

tan (x) = \frac{3}{3}

tan (x) = \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}

x = \frac{π}{6} + πn

x = \frac{π}{6} + πn

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

合并所有解

x = \frac{π}{6} + πn, x = πn

作图

流行的例子

4 sin (x) + 6 = 6

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

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

sin(θ)= 2/(sqrt(13)),cos(θ)= 3/(sqrt(13))

sin (θ) = \frac{2}{13}, cos (θ) = \frac{3}{13}

6sin^2(x)+4sin(x)-1=0

6 sin^{2} (x) + 4 sin (x) - 1 = 0

cos(x-pi/4)+sin(x+pi/4)=0

cos (x - \frac{π}{4}) + sin (x + \frac{π}{4}) = 0