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受欢迎的三角函数 >

3tan^2(θ)+sqrt(3)tan(θ)=0,0<= θ<= 360

解答

3 tan^{2} (θ) + 3 tan (θ) = 0, 0^{\circ} \leq θ \leq 36 0^{\circ}

解答

θ = 0, θ = 18 0^{\circ}, θ = 36 0^{\circ}, θ = 15 0^{\circ}, θ = 33 0^{\circ}

+1

弧度

θ = 0, θ = π, θ = 2 π, θ = \frac{5 π}{6}, θ = \frac{11 π}{6}

求解步骤

3 tan^{2} (θ) + 3 tan (θ) = 0, 0^{\circ} \leq θ \leq 36 0^{\circ}

用替代法求解

3 tan^{2} (θ) + 3 tan (θ) = 0

令

: tan (θ) = u

3 u^{2} + 3 u = 0

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

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} : 0

\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

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

\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}

使用分式法则:

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

= - \frac{2 3}{6}

约分:

2

= - \frac{3}{3}

二次方程组的解是:

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

u = tan (θ)

代回

tan (θ) = 0, tan (θ) = - \frac{3}{3}

tan (θ) = 0, tan (θ) = - \frac{3}{3}

tan (θ) = 0, 0 \leq θ \leq 36 0^{\circ} : θ = 0, θ = 18 0^{\circ}, θ = 36 0^{\circ}

tan (θ) = 0, 0 \leq θ \leq 36 0^{\circ}

tan (θ) = 0

的通解

tan (x)

周期表（周期为

18 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = 0 + 18 0^{\circ} n

θ = 0 + 18 0^{\circ} n

解

θ = 0 + 18 0^{\circ} n : θ = 18 0^{\circ} n

θ = 0 + 18 0^{\circ} n

0 + 18 0^{\circ} n = 18 0^{\circ} n

θ = 18 0^{\circ} n

θ = 18 0^{\circ} n

在

0 \leq θ \leq 36 0^{\circ}

范围内的解

θ = 0, θ = 18 0^{\circ}, θ = 36 0^{\circ}

tan (θ) = - \frac{3}{3}, 0 \leq θ \leq 36 0^{\circ} : θ = 15 0^{\circ}, θ = 33 0^{\circ}

tan (θ) = - \frac{3}{3}, 0 \leq θ \leq 36 0^{\circ}

tan (θ) = - \frac{3}{3}

的通解

tan (x)

周期表（周期为

18 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = 15 0^{\circ} + 18 0^{\circ} n

θ = 15 0^{\circ} + 18 0^{\circ} n

在

0 \leq θ \leq 36 0^{\circ}

范围内的解

θ = 15 0^{\circ}, θ = 33 0^{\circ}

合并所有解

θ = 0, θ = 18 0^{\circ}, θ = 36 0^{\circ}, θ = 15 0^{\circ}, θ = 33 0^{\circ}

作图

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