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)+4tan(x)=6

解答

sec^{2} (x) + 4 tan (x) = 6

解答

x = \frac{π}{4} + πn, x = - 1.37340 \dots + πn

+1

度数

x = 4 5^{\circ} + 18 0^{\circ} n, x = - 78.69006 \dots^{\circ} + 18 0^{\circ} n

求解步骤

sec^{2} (x) + 4 tan (x) = 6

两边减去

6

sec^{2} (x) + 4 tan (x) - 6 = 0

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

= - 6 + tan^{2} (x) + 1 + 4 tan (x)

化简

- 6 + tan^{2} (x) + 1 + 4 tan (x) : tan^{2} (x) + 4 tan (x) - 5

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

对同类项分组

= tan^{2} (x) + 4 tan (x) - 6 + 1

数字相加/相减:

- 6 + 1 = - 5

= tan^{2} (x) + 4 tan (x) - 5

= tan^{2} (x) + 4 tan (x) - 5

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

用替代法求解

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

令

: tan (x) = u

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

- 5 + u^{2} + 4 u = 0 : u = 1, u = - 5

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = 1, b = 4, c = - 5

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

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

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

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

使用法则

- (- a) = a

= 4^{2} + 4 \cdot 1 \cdot 5

数字相乘:

4 \cdot 1 \cdot 5 = 20

= 4^{2} + 20

4^{2} = 16

= 16 + 20

数字相加:

16 + 20 = 36

= 36

因式分解数字:

36 = 6^{2}

= 6^{2}

使用根式运算法则:

n a^{n} = a

6^{2} = 6

= 6

u_{1, 2} = \frac{- 4 \pm 6}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- 4 + 6}{2 \cdot 1}, u_{2} = \frac{- 4 - 6}{2 \cdot 1}

u = \frac{- 4 + 6}{2 \cdot 1} : 1

\frac{- 4 + 6}{2 \cdot 1}

数字相加/相减:

- 4 + 6 = 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{- 4 - 6}{2 \cdot 1} : - 5

\frac{- 4 - 6}{2 \cdot 1}

数字相减:

- 4 - 6 = - 10

= \frac{- 10}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 10}{2}

使用分式法则:

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

= - \frac{10}{2}

数字相除:

\frac{10}{2} = 5

= - 5

二次方程组的解是:

u = 1, u = - 5

u = tan (x)

代回

tan (x) = 1, tan (x) = - 5

tan (x) = 1, tan (x) = - 5

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) = - 5 : x = arctan (- 5) + πn

tan (x) = - 5

使用反三角函数性质

tan (x) = - 5

tan (x) = - 5

的通解

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

x = arctan (- 5) + πn

x = arctan (- 5) + πn

合并所有解

x = \frac{π}{4} + πn, x = arctan (- 5) + πn

以小数形式表示解

x = \frac{π}{4} + πn, x = - 1.37340 \dots + πn

作图

流行的例子

sin^2(θ)=4-2cos^2(θ)

sin^{2} (θ) = 4 - 2 cos^{2} (θ)

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

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

arcsin(x)= pi/7

arcsin (x) = \frac{π}{7}

2 sin (2 θ) = - 1

sin(4θ)=-(2sqrt(3))/3

sin (4 θ) = - \frac{2 3}{3}