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受欢迎的三角函数 >

tan^2(x)+sec(x)=5

解答

tan^{2} (x) + sec (x) = 5

解答

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = 1.91063 \dots + 2 πn, x = - 1.91063 \dots + 2 πn

+1

度数

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n, x = 109.47122 \dots^{\circ} + 36 0^{\circ} n, x = - 109.47122 \dots^{\circ} + 36 0^{\circ} n

求解步骤

tan^{2} (x) + sec (x) = 5

两边减去

5

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

使用三角恒等式改写

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

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

化简

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

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

对同类项分组

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

数字相减:

- 5 - 1 = - 6

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

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

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

用替代法求解

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

令

: sec (x) = u

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

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

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

改写成标准形式

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

u^{2} + u - 6 = 0

使用求根公式求解

u^{2} + u - 6 = 0

二次方程求根公式:

若

a = 1, b = 1, c = - 6

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

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

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

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 6)

使用法则

- (- a) = a

= 1 + 4 \cdot 1 \cdot 6

数字相乘:

4 \cdot 1 \cdot 6 = 24

= 1 + 24

数字相加:

1 + 24 = 25

= 25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- 1 + 5}{2 \cdot 1}, u_{2} = \frac{- 1 - 5}{2 \cdot 1}

u = \frac{- 1 + 5}{2 \cdot 1} : 2

\frac{- 1 + 5}{2 \cdot 1}

数字相加/相减:

- 1 + 5 = 4

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

数字相乘:

2 \cdot 1 = 2

= \frac{4}{2}

数字相除:

\frac{4}{2} = 2

= 2

u = \frac{- 1 - 5}{2 \cdot 1} : - 3

\frac{- 1 - 5}{2 \cdot 1}

数字相减:

- 1 - 5 = - 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 = 2, u = - 3

u = sec (x)

代回

sec (x) = 2, sec (x) = - 3

sec (x) = 2, sec (x) = - 3

sec (x) = 2 : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

sec (x) = 2

sec (x) = 2

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 = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

sec (x) = - 3 : x = arcsec (- 3) + 2 πn, x = - arcsec (- 3) + 2 πn

sec (x) = - 3

使用反三角函数性质

sec (x) = - 3

sec (x) = - 3

的通解

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

x = arcsec (- 3) + 2 πn, x = - arcsec (- 3) + 2 πn

x = arcsec (- 3) + 2 πn, x = - arcsec (- 3) + 2 πn

合并所有解

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = arcsec (- 3) + 2 πn, x = - arcsec (- 3) + 2 πn

以小数形式表示解

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = 1.91063 \dots + 2 πn, x = - 1.91063 \dots + 2 πn

作图

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