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 >

3sec^2(x)-2tan(x)-4=0

解答

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

解答

x = \frac{π}{4} + πn, x = - 0.32175 \dots + πn

+1

度数

x = 4 5^{\circ} + 18 0^{\circ} n, x = - 18.43494 \dots^{\circ} + 18 0^{\circ} n

求解步骤

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

化简

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

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

乘开

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

3 (tan^{2} (x) + 1)

使用分配律:

a (b + c) = ab + a c

a = 3, b = tan^{2} (x), c = 1

= 3 tan^{2} (x) + 3 \cdot 1

数字相乘:

3 \cdot 1 = 3

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

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

化简

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

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

对同类项分组

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

数字相加/相减:

- 4 + 3 = - 1

= 3 tan^{2} (x) - 2 tan (x) - 1

= 3 tan^{2} (x) - 2 tan (x) - 1

= 3 tan^{2} (x) - 2 tan (x) - 1

- 1 - 2 tan (x) + 3 tan^{2} (x) = 0

用替代法求解

- 1 - 2 tan (x) + 3 tan^{2} (x) = 0

令

: tan (x) = u

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

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

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

改写成标准形式

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

3 u^{2} - 2 u - 1 = 0

使用求根公式求解

3 u^{2} - 2 u - 1 = 0

二次方程求根公式:

若

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

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

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

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

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

使用法则

- (- a) = a

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

使用指数法则:

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

若

n

是偶数

(- 2)^{2} = 2^{2}

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

数字相乘:

4 \cdot 3 \cdot 1 = 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 3}

将解分隔开

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

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

\frac{- ( - 2 ) + 4}{2 \cdot 3}

使用法则

- (- a) = a

= \frac{2 + 4}{2 \cdot 3}

数字相加:

2 + 4 = 6

= \frac{6}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{6}{6}

使用法则

\frac{a}{a} = 1

= 1

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

\frac{- ( - 2 ) - 4}{2 \cdot 3}

使用法则

- (- a) = a

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

数字相减:

2 - 4 = - 2

= \frac{- 2}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{- 2}{6}

使用分式法则:

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

= - \frac{2}{6}

约分:

2

= - \frac{1}{3}

二次方程组的解是:

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

u = tan (x)

代回

tan (x) = 1, tan (x) = - \frac{1}{3}

tan (x) = 1, tan (x) = - \frac{1}{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) = - \frac{1}{3} : x = arctan (- \frac{1}{3}) + πn

tan (x) = - \frac{1}{3}

使用反三角函数性质

tan (x) = - \frac{1}{3}

tan (x) = - \frac{1}{3}

的通解

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

x = arctan (- \frac{1}{3}) + πn

x = arctan (- \frac{1}{3}) + πn

合并所有解

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

以小数形式表示解

x = \frac{π}{4} + πn, x = - 0.32175 \dots + πn

作图

流行的例子

9 sin^{2} (x) - 1 = 0

sin(x)= 1/(3.5)

sin (x) = \frac{1}{3.5}

tan(θ/3)=3sin(48)

tan (\frac{θ}{3}) = 3 sin (4 8^{\circ})

(cos(x)cot(x))/(1-sin(x))=csc(x)

\frac{cos ( x ) cot ( x )}{1 - sin ( x )} = csc (x)

2cos^2(x)-9sin(x)-3=0

2 cos^{2} (x) - 9 sin (x) - 3 = 0