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 >

tan^2(x)+4tan(x)=3

解答

tan^{2} (x) + 4 tan (x) = 3

解答

x = 0.57338 \dots + πn, x = - 1.35878 \dots + πn

+1

度数

x = 32.85240 \dots^{\circ} + 18 0^{\circ} n, x = - 77.85240 \dots^{\circ} + 18 0^{\circ} n

求解步骤

tan^{2} (x) + 4 tan (x) = 3

用替代法求解

tan^{2} (x) + 4 tan (x) = 3

令

: tan (x) = u

u^{2} + 4 u = 3

u^{2} + 4 u = 3 : u = - 2 + 7, u = - 2 - 7

u^{2} + 4 u = 3

将

3

para o lado esquerdo

u^{2} + 4 u = 3

两边减去

3

u^{2} + 4 u - 3 = 3 - 3

化简

u^{2} + 4 u - 3 = 0

u^{2} + 4 u - 3 = 0

使用求根公式求解

u^{2} + 4 u - 3 = 0

二次方程求根公式:

若

a = 1, b = 4, c = - 3

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

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

4^{2} - 4 \cdot 1 \cdot (- 3) = 27

4^{2} - 4 \cdot 1 \cdot (- 3)

使用法则

- (- a) = a

= 4^{2} + 4 \cdot 1 \cdot 3

数字相乘:

4 \cdot 1 \cdot 3 = 12

= 4^{2} + 12

4^{2} = 16

= 16 + 12

数字相加:

16 + 12 = 28

= 28

28

质因数分解:

2^{2} \cdot 7

28

28

除以

228 = 14 \cdot 2

= 2 \cdot 14

14

除以

214 = 7 \cdot 2

= 2 \cdot 2 \cdot 7

2, 7

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 7

= 2^{2} \cdot 7

= 2^{2} \cdot 7

使用根式运算法则:

n ab = n a n b

= 7 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 27

u_{1, 2} = \frac{- 4 \pm 2 7}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- 4 + 2 7}{2 \cdot 1}, u_{2} = \frac{- 4 - 2 7}{2 \cdot 1}

u = \frac{- 4 + 2 7}{2 \cdot 1} : - 2 + 7

\frac{- 4 + 2 7}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 4 + 2 7}{2}

分解

- 4 + 27 : 2 (- 2 + 7)

- 4 + 27

改写为

= - 2 \cdot 2 + 27

因式分解出通项

2

= 2 (- 2 + 7)

= \frac{2 ( - 2 + 7 )}{2}

数字相除:

\frac{2}{2} = 1

= - 2 + 7

u = \frac{- 4 - 2 7}{2 \cdot 1} : - 2 - 7

\frac{- 4 - 2 7}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 4 - 2 7}{2}

分解

- 4 - 27 : - 2 (2 + 7)

- 4 - 27

改写为

= - 2 \cdot 2 - 27

因式分解出通项

2

= - 2 (2 + 7)

= - \frac{2 ( 2 + 7 )}{2}

数字相除:

\frac{2}{2} = 1

= - (2 + 7)

取负

- (2 + 7) = - 2 - 7

= - 2 - 7

二次方程组的解是:

u = - 2 + 7, u = - 2 - 7

u = tan (x)

代回

tan (x) = - 2 + 7, tan (x) = - 2 - 7

tan (x) = - 2 + 7, tan (x) = - 2 - 7

tan (x) = - 2 + 7 : x = arctan (- 2 + 7) + πn

tan (x) = - 2 + 7

使用反三角函数性质

tan (x) = - 2 + 7

tan (x) = - 2 + 7

的通解

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

x = arctan (- 2 + 7) + πn

x = arctan (- 2 + 7) + πn

tan (x) = - 2 - 7 : x = arctan (- 2 - 7) + πn

tan (x) = - 2 - 7

使用反三角函数性质

tan (x) = - 2 - 7

tan (x) = - 2 - 7

的通解

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

x = arctan (- 2 - 7) + πn

x = arctan (- 2 - 7) + πn

合并所有解

x = arctan (- 2 + 7) + πn, x = arctan (- 2 - 7) + πn

以小数形式表示解

x = 0.57338 \dots + πn, x = - 1.35878 \dots + πn

作图

流行的例子

sin (1 - 2 x) = 1

5cos(x)+2=cos(x)

5 cos (x) + 2 = cos (x)

sec(x)=-5,pi<x<(3pi)/2

sec (x) = - 5, π < x < \frac{3 π}{2}

sin (\frac{θ}{2}) = \frac{3}{5}

2sin^3(t)+sin^2(t)-2sin(t)-1=0

2 sin^{3} (t) + sin^{2} (t) - 2 sin (t) - 1 = 0