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受欢迎的三角函数 >

2cos(x)-9=-4sec(x)

解答

2 cos (x) - 9 = - 4 sec (x)

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

2 cos (x) - 9 = - 4 sec (x)

两边减去

- 4 sec (x)

2 cos (x) - 9 + 4 sec (x) = 0

使用三角恒等式改写

- 9 + 2 cos (x) + 4 sec (x)

使用基本三角恒等式:

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

= - 9 + 2 \cdot \frac{1}{sec ( x )} + 4 sec (x)

2 \cdot \frac{1}{sec ( x )} = \frac{2}{sec ( x )}

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

分式相乘:

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

= \frac{1 \cdot 2}{sec ( x )}

数字相乘:

1 \cdot 2 = 2

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

= - 9 + \frac{2}{sec ( x )} + 4 sec (x)

- 9 + \frac{2}{sec ( x )} + 4 sec (x) = 0

用替代法求解

- 9 + \frac{2}{sec ( x )} + 4 sec (x) = 0

令

: sec (x) = u

- 9 + \frac{2}{u} + 4 u = 0

- 9 + \frac{2}{u} + 4 u = 0 : u = 2, u = \frac{1}{4}

- 9 + \frac{2}{u} + 4 u = 0

在两边乘以

u

- 9 + \frac{2}{u} + 4 u = 0

在两边乘以

u

- 9 u + \frac{2}{u} u + 4 uu = 0 \cdot u

化简

- 9 u + \frac{2}{u} u + 4 uu = 0 \cdot u

化简

\frac{2}{u} u : 2

\frac{2}{u} u

分式相乘:

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

= \frac{2 u}{u}

约分:

u

= 2

化简

4 uu : 4 u^{2}

4 uu

使用指数法则:

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

uu = u^{1 + 1}

= 4 u^{1 + 1}

数字相加:

1 + 1 = 2

= 4 u^{2}

化简

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

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

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

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

解

- 9 u + 2 + 4 u^{2} = 0 : u = 2, u = \frac{1}{4}

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

改写成标准形式

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

4 u^{2} - 9 u + 2 = 0

使用求根公式求解

4 u^{2} - 9 u + 2 = 0

二次方程求根公式:

若

a = 4, b = - 9, c = 2

u_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 4 \cdot 2}{2 \cdot 4}

u_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 4 \cdot 2}{2 \cdot 4}

(- 9)^{2} - 4 \cdot 4 \cdot 2 = 7

(- 9)^{2} - 4 \cdot 4 \cdot 2

使用指数法则:

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

若

n

是偶数

(- 9)^{2} = 9^{2}

= 9^{2} - 4 \cdot 4 \cdot 2

数字相乘:

4 \cdot 4 \cdot 2 = 32

= 9^{2} - 32

9^{2} = 81

= 81 - 32

数字相减:

81 - 32 = 49

= 49

因式分解数字:

49 = 7^{2}

= 7^{2}

使用根式运算法则:

n a^{n} = a

7^{2} = 7

= 7

u_{1, 2} = \frac{- ( - 9 ) \pm 7}{2 \cdot 4}

将解分隔开

u_{1} = \frac{- ( - 9 ) + 7}{2 \cdot 4}, u_{2} = \frac{- ( - 9 ) - 7}{2 \cdot 4}

u = \frac{- ( - 9 ) + 7}{2 \cdot 4} : 2

\frac{- ( - 9 ) + 7}{2 \cdot 4}

使用法则

- (- a) = a

= \frac{9 + 7}{2 \cdot 4}

数字相加:

9 + 7 = 16

= \frac{16}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{16}{8}

数字相除:

\frac{16}{8} = 2

= 2

u = \frac{- ( - 9 ) - 7}{2 \cdot 4} : \frac{1}{4}

\frac{- ( - 9 ) - 7}{2 \cdot 4}

使用法则

- (- a) = a

= \frac{9 - 7}{2 \cdot 4}

数字相减:

9 - 7 = 2

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

数字相乘:

2 \cdot 4 = 8

= \frac{2}{8}

约分:

2

= \frac{1}{4}

二次方程组的解是:

u = 2, u = \frac{1}{4}

u = 2, u = \frac{1}{4}

验证解

找到无定义的点（奇点）:

u = 0

取

- 9 + \frac{2}{u} + 4 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{1}{4}

u = sec (x)

代回

sec (x) = 2, sec (x) = \frac{1}{4}

sec (x) = 2, sec (x) = \frac{1}{4}

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{1}{4} :

无解

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

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

流行的例子

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4sin^2(x)+(2sqrt(3)+2)sin(x)+sqrt(3)=0

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3csc(x)=2csc(x)+sqrt(2)

3 csc (x) = 2 csc (x) + 2

sec(x)=tan(x)+cot(x)

sec (x) = tan (x) + cot (x)