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受欢迎的三角函数 >

5sec(x)-4cos(x)=8

解答

5 sec (x) - 4 cos (x) = 8

解答

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

+1

度数

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

求解步骤

5 sec (x) - 4 cos (x) = 8

两边减去

8

5 sec (x) - 4 cos (x) - 8 = 0

使用三角恒等式改写

- 8 - 4 cos (x) + 5 sec (x)

使用基本三角恒等式:

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

= - 8 - 4 \cdot \frac{1}{sec ( x )} + 5 sec (x)

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

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

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

数字相乘:

1 \cdot 4 = 4

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

= - 8 - \frac{4}{sec ( x )} + 5 sec (x)

- 8 - \frac{4}{sec ( x )} + 5 sec (x) = 0

用替代法求解

- 8 - \frac{4}{sec ( x )} + 5 sec (x) = 0

令

: sec (x) = u

- 8 - \frac{4}{u} + 5 u = 0

- 8 - \frac{4}{u} + 5 u = 0 : u = 2, u = - \frac{2}{5}

- 8 - \frac{4}{u} + 5 u = 0

在两边乘以

u

- 8 - \frac{4}{u} + 5 u = 0

在两边乘以

u

- 8 u - \frac{4}{u} u + 5 uu = 0 \cdot u

化简

- 8 u - \frac{4}{u} u + 5 uu = 0 \cdot u

化简

- \frac{4}{u} u : - 4

- \frac{4}{u} u

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= - \frac{4 u}{u}

约分:

u

= - 4

化简

5 uu : 5 u^{2}

5 uu

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

uu = u^{1 + 1}

= 5 u^{1 + 1}

数字相加:

1 + 1 = 2

= 5 u^{2}

化简

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

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

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

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

解

- 8 u - 4 + 5 u^{2} = 0 : u = 2, u = - \frac{2}{5}

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

改写成标准形式

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

5 u^{2} - 8 u - 4 = 0

使用求根公式求解

5 u^{2} - 8 u - 4 = 0

二次方程求根公式:

若

a = 5, b = - 8, c = - 4

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 5 ( - 4 )}{2 \cdot 5}

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 5 ( - 4 )}{2 \cdot 5}

(- 8)^{2} - 4 \cdot 5 (- 4) = 12

(- 8)^{2} - 4 \cdot 5 (- 4)

使用法则

- (- a) = a

= (- 8)^{2} + 4 \cdot 5 \cdot 4

使用指数法则:

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

若

n

是偶数

(- 8)^{2} = 8^{2}

= 8^{2} + 4 \cdot 5 \cdot 4

数字相乘:

4 \cdot 5 \cdot 4 = 80

= 8^{2} + 80

8^{2} = 64

= 64 + 80

数字相加:

64 + 80 = 144

= 144

因式分解数字:

144 = 1 2^{2}

= 1 2^{2}

使用根式运算法则:

n a^{n} = a

1 2^{2} = 12

= 12

u_{1, 2} = \frac{- ( - 8 ) \pm 12}{2 \cdot 5}

将解分隔开

u_{1} = \frac{- ( - 8 ) + 12}{2 \cdot 5}, u_{2} = \frac{- ( - 8 ) - 12}{2 \cdot 5}

u = \frac{- ( - 8 ) + 12}{2 \cdot 5} : 2

\frac{- ( - 8 ) + 12}{2 \cdot 5}

使用法则

- (- a) = a

= \frac{8 + 12}{2 \cdot 5}

数字相加:

8 + 12 = 20

= \frac{20}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{20}{10}

数字相除:

\frac{20}{10} = 2

= 2

u = \frac{- ( - 8 ) - 12}{2 \cdot 5} : - \frac{2}{5}

\frac{- ( - 8 ) - 12}{2 \cdot 5}

使用法则

- (- a) = a

= \frac{8 - 12}{2 \cdot 5}

数字相减:

8 - 12 = - 4

= \frac{- 4}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{- 4}{10}

使用分式法则:

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

= - \frac{4}{10}

约分:

2

= - \frac{2}{5}

二次方程组的解是:

u = 2, u = - \frac{2}{5}

u = 2, u = - \frac{2}{5}

验证解

找到无定义的点（奇点）:

u = 0

取

- 8 - \frac{4}{u} + 5 u

的分母，令其等于零

u = 0

以下点无定义

u = 0

将不在定义域的点与解相综合:

u = 2, u = - \frac{2}{5}

u = sec (x)

代回

sec (x) = 2, sec (x) = - \frac{2}{5}

sec (x) = 2, sec (x) = - \frac{2}{5}

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) = - \frac{2}{5} :

无解

sec (x) = - \frac{2}{5}

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

无解

合并所有解

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

作图

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