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 >

4tan^2(θ)+tan(θ)=-4tan(θ)-1

解答

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

解答

θ = - 0.24497 \dots + πn, θ = \frac{3 π}{4} + πn

+1

度数

θ = - 14.03624 \dots^{\circ} + 18 0^{\circ} n, θ = 13 5^{\circ} + 18 0^{\circ} n

求解步骤

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

用替代法求解

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

令

: tan (θ) = u

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

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

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

将

1

para o lado esquerdo

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

两边加上

1

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

化简

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

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

将

4 u

para o lado esquerdo

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

两边加上

4 u

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

化简

4 u^{2} + 5 u + 1 = 0

4 u^{2} + 5 u + 1 = 0

使用求根公式求解

4 u^{2} + 5 u + 1 = 0

二次方程求根公式:

若

a = 4, b = 5, c = 1

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 4 \cdot 1}{2 \cdot 4}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 4 \cdot 1}{2 \cdot 4}

5^{2} - 4 \cdot 4 \cdot 1 = 3

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

数字相乘:

4 \cdot 4 \cdot 1 = 16

= 5^{2} - 16

5^{2} = 25

= 25 - 16

数字相减:

25 - 16 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 3

u_{1, 2} = \frac{- 5 \pm 3}{2 \cdot 4}

将解分隔开

u_{1} = \frac{- 5 + 3}{2 \cdot 4}, u_{2} = \frac{- 5 - 3}{2 \cdot 4}

u = \frac{- 5 + 3}{2 \cdot 4} : - \frac{1}{4}

\frac{- 5 + 3}{2 \cdot 4}

数字相加/相减:

- 5 + 3 = - 2

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

数字相乘:

2 \cdot 4 = 8

= \frac{- 2}{8}

使用分式法则:

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

= - \frac{2}{8}

约分:

2

= - \frac{1}{4}

u = \frac{- 5 - 3}{2 \cdot 4} : - 1

\frac{- 5 - 3}{2 \cdot 4}

数字相减:

- 5 - 3 = - 8

= \frac{- 8}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 8}{8}

使用分式法则:

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

= - \frac{8}{8}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

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

u = tan (θ)

代回

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

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

tan (θ) = - \frac{1}{4} : θ = arctan (- \frac{1}{4}) + πn

tan (θ) = - \frac{1}{4}

使用反三角函数性质

tan (θ) = - \frac{1}{4}

tan (θ) = - \frac{1}{4}

的通解

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

θ = arctan (- \frac{1}{4}) + πn

θ = arctan (- \frac{1}{4}) + πn

tan (θ) = - 1 : θ = \frac{3 π}{4} + πn

tan (θ) = - 1

tan (θ) = - 1

的通解

tan (x)

周期表（周期为

πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

合并所有解

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

以小数形式表示解

θ = - 0.24497 \dots + πn, θ = \frac{3 π}{4} + πn

作图

流行的例子

8cos(c)+6=3cos(c)+7

8 cos (c) + 6 = 3 cos (c) + 7

2 csc (A) - 4 = 0

7 tan^{2} (x) - 21 = 0

-4cos(x)=-sin^2(x)+4

- 4 cos (x) = - sin^{2} (x) + 4

sin(x)cos(x)+sin(x)=0

sin (x) cos (x) + sin (x) = 0