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受欢迎的三角函数 >

6tan^2(θ)+10tan(θ)=9tan(θ)+2

解答

6 tan^{2} (θ) + 10 tan (θ) = 9 tan (θ) + 2

解答

θ = 0.46364 \dots + πn, θ = - 0.58800 \dots + πn

+1

度数

θ = 26.56505 \dots^{\circ} + 18 0^{\circ} n, θ = - 33.69006 \dots^{\circ} + 18 0^{\circ} n

求解步骤

6 tan^{2} (θ) + 10 tan (θ) = 9 tan (θ) + 2

用替代法求解

6 tan^{2} (θ) + 10 tan (θ) = 9 tan (θ) + 2

令

: tan (θ) = u

6 u^{2} + 10 u = 9 u + 2

6 u^{2} + 10 u = 9 u + 2 : u = \frac{1}{2}, u = - \frac{2}{3}

6 u^{2} + 10 u = 9 u + 2

将

2

para o lado esquerdo

6 u^{2} + 10 u = 9 u + 2

两边减去

2

6 u^{2} + 10 u - 2 = 9 u + 2 - 2

化简

6 u^{2} + 10 u - 2 = 9 u

6 u^{2} + 10 u - 2 = 9 u

将

9 u

para o lado esquerdo

6 u^{2} + 10 u - 2 = 9 u

两边减去

9 u

6 u^{2} + 10 u - 2 - 9 u = 9 u - 9 u

化简

6 u^{2} + u - 2 = 0

6 u^{2} + u - 2 = 0

使用求根公式求解

6 u^{2} + u - 2 = 0

二次方程求根公式:

若

a = 6, b = 1, c = - 2

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 6 ( - 2 )}{2 \cdot 6}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 6 ( - 2 )}{2 \cdot 6}

1^{2} - 4 \cdot 6 (- 2) = 7

1^{2} - 4 \cdot 6 (- 2)

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 6 (- 2)

使用法则

- (- a) = a

= 1 + 4 \cdot 6 \cdot 2

数字相乘:

4 \cdot 6 \cdot 2 = 48

= 1 + 48

数字相加:

1 + 48 = 49

= 49

因式分解数字:

49 = 7^{2}

= 7^{2}

使用根式运算法则:

n a^{n} = a

7^{2} = 7

= 7

u_{1, 2} = \frac{- 1 \pm 7}{2 \cdot 6}

将解分隔开

u_{1} = \frac{- 1 + 7}{2 \cdot 6}, u_{2} = \frac{- 1 - 7}{2 \cdot 6}

u = \frac{- 1 + 7}{2 \cdot 6} : \frac{1}{2}

\frac{- 1 + 7}{2 \cdot 6}

数字相加/相减:

- 1 + 7 = 6

= \frac{6}{2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{6}{12}

约分:

6

= \frac{1}{2}

u = \frac{- 1 - 7}{2 \cdot 6} : - \frac{2}{3}

\frac{- 1 - 7}{2 \cdot 6}

数字相减:

- 1 - 7 = - 8

= \frac{- 8}{2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{- 8}{12}

使用分式法则:

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

= - \frac{8}{12}

约分:

4

= - \frac{2}{3}

二次方程组的解是:

u = \frac{1}{2}, u = - \frac{2}{3}

u = tan (θ)

代回

tan (θ) = \frac{1}{2}, tan (θ) = - \frac{2}{3}

tan (θ) = \frac{1}{2}, tan (θ) = - \frac{2}{3}

tan (θ) = \frac{1}{2} : θ = arctan (\frac{1}{2}) + πn

tan (θ) = \frac{1}{2}

使用反三角函数性质

tan (θ) = \frac{1}{2}

tan (θ) = \frac{1}{2}

的通解

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

θ = arctan (\frac{1}{2}) + πn

θ = arctan (\frac{1}{2}) + πn

tan (θ) = - \frac{2}{3} : θ = arctan (- \frac{2}{3}) + πn

tan (θ) = - \frac{2}{3}

使用反三角函数性质

tan (θ) = - \frac{2}{3}

tan (θ) = - \frac{2}{3}

的通解

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

θ = arctan (- \frac{2}{3}) + πn

θ = arctan (- \frac{2}{3}) + πn

合并所有解

θ = arctan (\frac{1}{2}) + πn, θ = arctan (- \frac{2}{3}) + πn

以小数形式表示解

θ = 0.46364 \dots + πn, θ = - 0.58800 \dots + πn

作图

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