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受欢迎的三角函数 >

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

解答

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

解答

x = 1.36943 \dots + 2 πn, x = 2 π - 1.36943 \dots + 2 πn, x = π + 2 πn

+1

度数

x = 78.46304 \dots^{\circ} + 36 0^{\circ} n, x = 281.53695 \dots^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

4

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

化简

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

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

对同类项分组

= sec^{2} (x) - 4 sec (x) - 4 - 1

数字相减:

- 4 - 1 = - 5

= sec^{2} (x) - 4 sec (x) - 5

= sec^{2} (x) - 4 sec (x) - 5

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

用替代法求解

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

令

: sec (x) = u

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

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

- 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

使用指数法则:

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

若

n

是偶数

(- 4)^{2} = 4^{2}

= 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} : 5

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

使用法则

- (- a) = a

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

数字相加:

4 + 6 = 10

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

数字相乘:

2 \cdot 1 = 2

= \frac{10}{2}

数字相除:

\frac{10}{2} = 5

= 5

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

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

使用法则

- (- a) = a

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

数字相减:

4 - 6 = - 2

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

数字相乘:

2 \cdot 1 = 2

= \frac{- 2}{2}

使用分式法则:

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

= - \frac{2}{2}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

u = 5, u = - 1

u = sec (x)

代回

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

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

sec (x) = 5 : x = arcsec (5) + 2 πn, x = 2 π - arcsec (5) + 2 πn

sec (x) = 5

使用反三角函数性质

sec (x) = 5

sec (x) = 5

的通解

sec (x) = a \Rightarrow x = arcsec (a) + 2 πn, x = 2 π - arcsec (a) + 2 πn

x = arcsec (5) + 2 πn, x = 2 π - arcsec (5) + 2 πn

x = arcsec (5) + 2 πn, x = 2 π - arcsec (5) + 2 πn

sec (x) = - 1 : x = π + 2 πn

sec (x) = - 1

sec (x) = - 1

的通解

sec (x)

周期表（周期为

2 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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

x = π + 2 πn

x = π + 2 πn

合并所有解

x = arcsec (5) + 2 πn, x = 2 π - arcsec (5) + 2 πn, x = π + 2 πn

以小数形式表示解

x = 1.36943 \dots + 2 πn, x = 2 π - 1.36943 \dots + 2 πn, x = π + 2 πn

作图

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