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 >

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

解答

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

解答

x = \frac{π}{4} + πn, x = - 1.24904 \dots + πn

+1

度数

x = 4 5^{\circ} + 18 0^{\circ} n, x = - 71.56505 \dots^{\circ} + 18 0^{\circ} n

求解步骤

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

两边减去

4

sec^{2} (x) + 2 tan (x) - 4 = 0

使用三角恒等式改写

- 4 + sec^{2} (x) + 2 tan (x)

使用毕达哥拉斯恒等式:

sec^{2} (x) = tan^{2} (x) + 1

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

化简

- 4 + tan^{2} (x) + 1 + 2 tan (x) : tan^{2} (x) + 2 tan (x) - 3

- 4 + tan^{2} (x) + 1 + 2 tan (x)

对同类项分组

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

数字相加/相减:

- 4 + 1 = - 3

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

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

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

用替代法求解

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

令

: tan (x) = u

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

- 3 + u^{2} + 2 u = 0 : u = 1, u = - 3

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

改写成标准形式

a x^{2} + b x + c = 0

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

使用求根公式求解

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

二次方程求根公式:

若

a = 1, b = 2, c = - 3

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

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

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

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

使用法则

- (- a) = a

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

数字相乘:

4 \cdot 1 \cdot 3 = 12

= 2^{2} + 12

2^{2} = 4

= 4 + 12

数字相加:

4 + 12 = 16

= 16

因式分解数字:

16 = 4^{2}

= 4^{2}

使用根式运算法则:

n a^{n} = a

4^{2} = 4

= 4

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

将解分隔开

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

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

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

数字相加/相减:

- 2 + 4 = 2

= \frac{2}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{2}{2}

使用法则

\frac{a}{a} = 1

= 1

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

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

数字相减:

- 2 - 4 = - 6

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

数字相乘:

2 \cdot 1 = 2

= \frac{- 6}{2}

使用分式法则:

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

= - \frac{6}{2}

数字相除:

\frac{6}{2} = 3

= - 3

二次方程组的解是:

u = 1, u = - 3

u = tan (x)

代回

tan (x) = 1, tan (x) = - 3

tan (x) = 1, tan (x) = - 3

tan (x) = 1 : x = \frac{π}{4} + πn

tan (x) = 1

tan (x) = 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}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

tan (x) = - 3 : x = arctan (- 3) + πn

tan (x) = - 3

使用反三角函数性质

tan (x) = - 3

tan (x) = - 3

的通解

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

x = arctan (- 3) + πn

x = arctan (- 3) + πn

合并所有解

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

以小数形式表示解

x = \frac{π}{4} + πn, x = - 1.24904 \dots + πn

作图

流行的例子

tan(x/5)+sqrt(3)=0

tan (\frac{x}{5}) + 3 = 0

cos(2x)-sqrt(2)cos(x)=-1

cos (2 x) - 2 cos (x) = - 1

2sin(x+pi/2)+3tan(pi-x)=0

2 sin (x + \frac{π}{2}) + 3 tan (π - x) = 0

sin (x) = 0.27

sec(x/2)=cos(x/2)

sec (\frac{x}{2}) = cos (\frac{x}{2})