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 >

-8tan^2(θ)-4tan(θ)-11=6tan(θ)-8

解答

- 8 tan^{2} (θ) - 4 tan (θ) - 11 = 6 tan (θ) - 8

解答

θ = - 0.64350 \dots + πn, θ = - 0.46364 \dots + πn

+1

度数

θ = - 36.86989 \dots^{\circ} + 18 0^{\circ} n, θ = - 26.56505 \dots^{\circ} + 18 0^{\circ} n

求解步骤

- 8 tan^{2} (θ) - 4 tan (θ) - 11 = 6 tan (θ) - 8

用替代法求解

- 8 tan^{2} (θ) - 4 tan (θ) - 11 = 6 tan (θ) - 8

令

: tan (θ) = u

- 8 u^{2} - 4 u - 11 = 6 u - 8

- 8 u^{2} - 4 u - 11 = 6 u - 8 : u = - \frac{3}{4}, u = - \frac{1}{2}

- 8 u^{2} - 4 u - 11 = 6 u - 8

将

8

para o lado esquerdo

- 8 u^{2} - 4 u - 11 = 6 u - 8

两边加上

8

- 8 u^{2} - 4 u - 11 + 8 = 6 u - 8 + 8

化简

- 8 u^{2} - 4 u - 3 = 6 u

- 8 u^{2} - 4 u - 3 = 6 u

将

6 u

para o lado esquerdo

- 8 u^{2} - 4 u - 3 = 6 u

两边减去

6 u

- 8 u^{2} - 4 u - 3 - 6 u = 6 u - 6 u

化简

- 8 u^{2} - 10 u - 3 = 0

- 8 u^{2} - 10 u - 3 = 0

使用求根公式求解

- 8 u^{2} - 10 u - 3 = 0

二次方程求根公式:

若

a = - 8, b = - 10, c = - 3

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 ( - 8 ) ( - 3 )}{2 ( - 8 )}

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 ( - 8 ) ( - 3 )}{2 ( - 8 )}

(- 10)^{2} - 4 (- 8) (- 3) = 2

(- 10)^{2} - 4 (- 8) (- 3)

使用法则

- (- a) = a

= (- 10)^{2} - 4 \cdot 8 \cdot 3

使用指数法则:

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

若

n

是偶数

(- 10)^{2} = 1 0^{2}

= 1 0^{2} - 4 \cdot 8 \cdot 3

数字相乘:

4 \cdot 8 \cdot 3 = 96

= 1 0^{2} - 96

1 0^{2} = 100

= 100 - 96

数字相减:

100 - 96 = 4

= 4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

u_{1, 2} = \frac{- ( - 10 ) \pm 2}{2 ( - 8 )}

将解分隔开

u_{1} = \frac{- ( - 10 ) + 2}{2 ( - 8 )}, u_{2} = \frac{- ( - 10 ) - 2}{2 ( - 8 )}

u = \frac{- ( - 10 ) + 2}{2 ( - 8 )} : - \frac{3}{4}

\frac{- ( - 10 ) + 2}{2 ( - 8 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{10 + 2}{- 2 \cdot 8}

数字相加:

10 + 2 = 12

= \frac{12}{- 2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{12}{- 16}

使用分式法则:

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

= - \frac{12}{16}

约分:

4

= - \frac{3}{4}

u = \frac{- ( - 10 ) - 2}{2 ( - 8 )} : - \frac{1}{2}

\frac{- ( - 10 ) - 2}{2 ( - 8 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{10 - 2}{- 2 \cdot 8}

数字相减:

10 - 2 = 8

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

数字相乘:

2 \cdot 8 = 16

= \frac{8}{- 16}

使用分式法则:

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

= - \frac{8}{16}

约分:

8

= - \frac{1}{2}

二次方程组的解是:

u = - \frac{3}{4}, u = - \frac{1}{2}

u = tan (θ)

代回

tan (θ) = - \frac{3}{4}, tan (θ) = - \frac{1}{2}

tan (θ) = - \frac{3}{4}, tan (θ) = - \frac{1}{2}

tan (θ) = - \frac{3}{4} : θ = arctan (- \frac{3}{4}) + πn

tan (θ) = - \frac{3}{4}

使用反三角函数性质

tan (θ) = - \frac{3}{4}

tan (θ) = - \frac{3}{4}

的通解

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

θ = arctan (- \frac{3}{4}) + πn

θ = arctan (- \frac{3}{4}) + 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{3}{4}) + πn, θ = arctan (- \frac{1}{2}) + πn

以小数形式表示解

θ = - 0.64350 \dots + πn, θ = - 0.46364 \dots + πn

作图

流行的例子

2cos^2(x)+3cos(x)-5=0

2 cos^{2} (x) + 3 cos (x) - 5 = 0

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

1 + cos^{2} (x) = sin^{2} (x)

4 cos (2 x) = 2

cos (x) = \frac{5}{4}

sin^{2} (2 x) = \frac{1}{2}