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受欢迎的三角函数 >

4tan^2(x)+15sec(x)=0

解答

4 tan^{2} (x) + 15 sec (x) = 0

解答

x = 1.82347 \dots + 2 πn, x = - 1.82347 \dots + 2 πn

+1

度数

x = 104.47751 \dots^{\circ} + 36 0^{\circ} n, x = - 104.47751 \dots^{\circ} + 36 0^{\circ} n

求解步骤

4 tan^{2} (x) + 15 sec (x) = 0

使用三角恒等式改写

15 sec (x) + 4 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

= 15 sec (x) + 4 (sec^{2} (x) - 1)

(- 1 + sec^{2} (x)) \cdot 4 + 15 sec (x) = 0

用替代法求解

(- 1 + sec^{2} (x)) \cdot 4 + 15 sec (x) = 0

令

: sec (x) = u

(- 1 + u^{2}) \cdot 4 + 15 u = 0

(- 1 + u^{2}) \cdot 4 + 15 u = 0 : u = \frac{1}{4}, u = - 4

(- 1 + u^{2}) \cdot 4 + 15 u = 0

展开

(- 1 + u^{2}) \cdot 4 + 15 u : - 4 + 4 u^{2} + 15 u

(- 1 + u^{2}) \cdot 4 + 15 u

= 4 (- 1 + u^{2}) + 15 u

乘开

4 (- 1 + u^{2}) : - 4 + 4 u^{2}

4 (- 1 + u^{2})

使用分配律:

a (b + c) = ab + a c

a = 4, b = - 1, c = u^{2}

= 4 (- 1) + 4 u^{2}

使用加减运算法则

+ (- a) = - a

= - 4 \cdot 1 + 4 u^{2}

数字相乘:

4 \cdot 1 = 4

= - 4 + 4 u^{2}

= - 4 + 4 u^{2} + 15 u

- 4 + 4 u^{2} + 15 u = 0

改写成标准形式

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

4 u^{2} + 15 u - 4 = 0

使用求根公式求解

4 u^{2} + 15 u - 4 = 0

二次方程求根公式:

若

a = 4, b = 15, c = - 4

u_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 \cdot 4 ( - 4 )}{2 \cdot 4}

u_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 \cdot 4 ( - 4 )}{2 \cdot 4}

1 5^{2} - 4 \cdot 4 (- 4) = 17

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

使用法则

- (- a) = a

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

数字相乘:

4 \cdot 4 \cdot 4 = 64

= 1 5^{2} + 64

1 5^{2} = 225

= 225 + 64

数字相加:

225 + 64 = 289

= 289

因式分解数字:

289 = 1 7^{2}

= 1 7^{2}

使用根式运算法则:

n a^{n} = a

1 7^{2} = 17

= 17

u_{1, 2} = \frac{- 15 \pm 17}{2 \cdot 4}

将解分隔开

u_{1} = \frac{- 15 + 17}{2 \cdot 4}, u_{2} = \frac{- 15 - 17}{2 \cdot 4}

u = \frac{- 15 + 17}{2 \cdot 4} : \frac{1}{4}

\frac{- 15 + 17}{2 \cdot 4}

数字相加/相减:

- 15 + 17 = 2

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

数字相乘:

2 \cdot 4 = 8

= \frac{2}{8}

约分:

2

= \frac{1}{4}

u = \frac{- 15 - 17}{2 \cdot 4} : - 4

\frac{- 15 - 17}{2 \cdot 4}

数字相减:

- 15 - 17 = - 32

= \frac{- 32}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 32}{8}

使用分式法则:

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

= - \frac{32}{8}

数字相除:

\frac{32}{8} = 4

= - 4

二次方程组的解是:

u = \frac{1}{4}, u = - 4

u = sec (x)

代回

sec (x) = \frac{1}{4}, sec (x) = - 4

sec (x) = \frac{1}{4}, sec (x) = - 4

sec (x) = \frac{1}{4} :

无解

sec (x) = \frac{1}{4}

sec (x) \leq - 1 or sec (x) \geq 1

无解

sec (x) = - 4 : x = arcsec (- 4) + 2 πn, x = - arcsec (- 4) + 2 πn

sec (x) = - 4

使用反三角函数性质

sec (x) = - 4

sec (x) = - 4

的通解

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

x = arcsec (- 4) + 2 πn, x = - arcsec (- 4) + 2 πn

x = arcsec (- 4) + 2 πn, x = - arcsec (- 4) + 2 πn

合并所有解

x = arcsec (- 4) + 2 πn, x = - arcsec (- 4) + 2 πn

以小数形式表示解

x = 1.82347 \dots + 2 πn, x = - 1.82347 \dots + 2 πn

作图

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