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受欢迎的三角函数 >

-6tan^2(θ)+4tan(θ)=-tan(θ)+1

解答

- 6 tan^{2} (θ) + 4 tan (θ) = - tan (θ) + 1

解答

θ = 0.32175 \dots + πn, θ = 0.46364 \dots + πn

+1

度数

θ = 18.43494 \dots^{\circ} + 18 0^{\circ} n, θ = 26.56505 \dots^{\circ} + 18 0^{\circ} n

求解步骤

- 6 tan^{2} (θ) + 4 tan (θ) = - tan (θ) + 1

用替代法求解

- 6 tan^{2} (θ) + 4 tan (θ) = - tan (θ) + 1

令

: tan (θ) = u

- 6 u^{2} + 4 u = - u + 1

- 6 u^{2} + 4 u = - u + 1 : u = \frac{1}{3}, u = \frac{1}{2}

- 6 u^{2} + 4 u = - u + 1

将

1

para o lado esquerdo

- 6 u^{2} + 4 u = - u + 1

两边减去

1

- 6 u^{2} + 4 u - 1 = - u + 1 - 1

化简

- 6 u^{2} + 4 u - 1 = - u

- 6 u^{2} + 4 u - 1 = - u

将

u

para o lado esquerdo

- 6 u^{2} + 4 u - 1 = - u

两边加上

u

- 6 u^{2} + 4 u - 1 + u = - u + u

化简

- 6 u^{2} + 5 u - 1 = 0

- 6 u^{2} + 5 u - 1 = 0

使用求根公式求解

- 6 u^{2} + 5 u - 1 = 0

二次方程求根公式:

若

a = - 6, b = 5, c = - 1

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( - 6 ) ( - 1 )}{2 ( - 6 )}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( - 6 ) ( - 1 )}{2 ( - 6 )}

5^{2} - 4 (- 6) (- 1) = 1

5^{2} - 4 (- 6) (- 1)

使用法则

- (- a) = a

= 5^{2} - 4 \cdot 6 \cdot 1

数字相乘:

4 \cdot 6 \cdot 1 = 24

= 5^{2} - 24

5^{2} = 25

= 25 - 24

数字相减:

25 - 24 = 1

= 1

使用法则

1 = 1

= 1

u_{1, 2} = \frac{- 5 \pm 1}{2 ( - 6 )}

将解分隔开

u_{1} = \frac{- 5 + 1}{2 ( - 6 )}, u_{2} = \frac{- 5 - 1}{2 ( - 6 )}

u = \frac{- 5 + 1}{2 ( - 6 )} : \frac{1}{3}

\frac{- 5 + 1}{2 ( - 6 )}

去除括号:

(- a) = - a

= \frac{- 5 + 1}{- 2 \cdot 6}

数字相加/相减:

- 5 + 1 = - 4

= \frac{- 4}{- 2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{- 4}{- 12}

使用分式法则:

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

= \frac{4}{12}

约分:

4

= \frac{1}{3}

u = \frac{- 5 - 1}{2 ( - 6 )} : \frac{1}{2}

\frac{- 5 - 1}{2 ( - 6 )}

去除括号:

(- a) = - a

= \frac{- 5 - 1}{- 2 \cdot 6}

数字相减:

- 5 - 1 = - 6

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

数字相乘:

2 \cdot 6 = 12

= \frac{- 6}{- 12}

使用分式法则:

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

= \frac{6}{12}

约分:

6

= \frac{1}{2}

二次方程组的解是:

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

u = tan (θ)

代回

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

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

tan (θ) = \frac{1}{3} : θ = arctan (\frac{1}{3}) + πn

tan (θ) = \frac{1}{3}

使用反三角函数性质

tan (θ) = \frac{1}{3}

tan (θ) = \frac{1}{3}

的通解

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

θ = arctan (\frac{1}{3}) + πn

θ = arctan (\frac{1}{3}) + πn

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

合并所有解

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

以小数形式表示解

θ = 0.32175 \dots + πn, θ = 0.46364 \dots + πn

作图

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