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受欢迎的三角函数 >

-6tan^2(θ)+11tan(θ)-6=-2tan(θ)

解答

- 6 tan^{2} (θ) + 11 tan (θ) - 6 = - 2 tan (θ)

解答

θ = 0.58800 \dots + πn, θ = 0.98279 \dots + πn

+1

度数

θ = 33.69006 \dots^{\circ} + 18 0^{\circ} n, θ = 56.30993 \dots^{\circ} + 18 0^{\circ} n

求解步骤

- 6 tan^{2} (θ) + 11 tan (θ) - 6 = - 2 tan (θ)

用替代法求解

- 6 tan^{2} (θ) + 11 tan (θ) - 6 = - 2 tan (θ)

令

: tan (θ) = u

- 6 u^{2} + 11 u - 6 = - 2 u

- 6 u^{2} + 11 u - 6 = - 2 u : u = \frac{2}{3}, u = \frac{3}{2}

- 6 u^{2} + 11 u - 6 = - 2 u

将

2 u

para o lado esquerdo

- 6 u^{2} + 11 u - 6 = - 2 u

两边加上

2 u

- 6 u^{2} + 11 u - 6 + 2 u = - 2 u + 2 u

化简

- 6 u^{2} + 13 u - 6 = 0

- 6 u^{2} + 13 u - 6 = 0

使用求根公式求解

- 6 u^{2} + 13 u - 6 = 0

二次方程求根公式:

若

a = - 6, b = 13, c = - 6

u_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 ( - 6 ) ( - 6 )}{2 ( - 6 )}

u_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 ( - 6 ) ( - 6 )}{2 ( - 6 )}

1 3^{2} - 4 (- 6) (- 6) = 5

1 3^{2} - 4 (- 6) (- 6)

使用法则

- (- a) = a

= 1 3^{2} - 4 \cdot 6 \cdot 6

数字相乘:

4 \cdot 6 \cdot 6 = 144

= 1 3^{2} - 144

1 3^{2} = 169

= 169 - 144

数字相减:

169 - 144 = 25

= 25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 5

u_{1, 2} = \frac{- 13 \pm 5}{2 ( - 6 )}

将解分隔开

u_{1} = \frac{- 13 + 5}{2 ( - 6 )}, u_{2} = \frac{- 13 - 5}{2 ( - 6 )}

u = \frac{- 13 + 5}{2 ( - 6 )} : \frac{2}{3}

\frac{- 13 + 5}{2 ( - 6 )}

去除括号:

(- a) = - a

= \frac{- 13 + 5}{- 2 \cdot 6}

数字相加/相减:

- 13 + 5 = - 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}

u = \frac{- 13 - 5}{2 ( - 6 )} : \frac{3}{2}

\frac{- 13 - 5}{2 ( - 6 )}

去除括号:

(- a) = - a

= \frac{- 13 - 5}{- 2 \cdot 6}

数字相减:

- 13 - 5 = - 18

= \frac{- 18}{- 2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{- 18}{- 12}

使用分式法则:

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

= \frac{18}{12}

约分:

6

= \frac{3}{2}

二次方程组的解是:

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

u = tan (θ)

代回

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

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

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

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

tan (θ) = \frac{3}{2}

使用反三角函数性质

tan (θ) = \frac{3}{2}

tan (θ) = \frac{3}{2}

的通解

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

θ = arctan (\frac{3}{2}) + πn

θ = arctan (\frac{3}{2}) + πn

合并所有解

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

以小数形式表示解

θ = 0.58800 \dots + πn, θ = 0.98279 \dots + πn

作图

流行的例子

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sin(θ)=-cos(θ)

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