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受欢迎的三角函数 >

cos(2x)+sec(2x)=11

解答

cos (2 x) + sec (2 x) = 11

解答

x = \frac{1.47899 \dots}{2} + πn, x = π - \frac{1.47899 \dots}{2} + πn

+1

度数

x = 42.37006 \dots^{\circ} + 18 0^{\circ} n, x = 137.62993 \dots^{\circ} + 18 0^{\circ} n

求解步骤

cos (2 x) + sec (2 x) = 11

两边减去

11

cos (2 x) + sec (2 x) - 11 = 0

使用三角恒等式改写

- 11 + cos (2 x) + sec (2 x)

使用基本三角恒等式:

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

= - 11 + \frac{1}{sec ( 2 x )} + sec (2 x)

- 11 + \frac{1}{sec ( 2 x )} + sec (2 x) = 0

用替代法求解

- 11 + \frac{1}{sec ( 2 x )} + sec (2 x) = 0

令

: sec (2 x) = u

- 11 + \frac{1}{u} + u = 0

- 11 + \frac{1}{u} + u = 0 : u = \frac{11 + 3 13}{2}, u = \frac{11 - 3 13}{2}

- 11 + \frac{1}{u} + u = 0

在两边乘以

u

- 11 + \frac{1}{u} + u = 0

在两边乘以

u

- 11 u + \frac{1}{u} u + uu = 0 \cdot u

化简

- 11 u + \frac{1}{u} u + uu = 0 \cdot u

化简

\frac{1}{u} u : 1

\frac{1}{u} u

分式相乘:

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

= \frac{1 \cdot u}{u}

约分:

u

= 1

化简

uu : u^{2}

uu

使用指数法则:

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

uu = u^{1 + 1}

= u^{1 + 1}

数字相加:

1 + 1 = 2

= u^{2}

化简

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

- 11 u + 1 + u^{2} = 0

- 11 u + 1 + u^{2} = 0

- 11 u + 1 + u^{2} = 0

解

- 11 u + 1 + u^{2} = 0 : u = \frac{11 + 3 13}{2}, u = \frac{11 - 3 13}{2}

- 11 u + 1 + u^{2} = 0

改写成标准形式

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

u^{2} - 11 u + 1 = 0

使用求根公式求解

u^{2} - 11 u + 1 = 0

二次方程求根公式:

若

a = 1, b = - 11, c = 1

u_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

(- 11)^{2} - 4 \cdot 1 \cdot 1 = 313

(- 11)^{2} - 4 \cdot 1 \cdot 1

使用指数法则:

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

若

n

是偶数

(- 11)^{2} = 1 1^{2}

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

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 1 1^{2} - 4

1 1^{2} = 121

= 121 - 4

数字相减:

121 - 4 = 117

= 117

117

质因数分解:

3^{2} \cdot 13

117

117

除以

3117 = 39 \cdot 3

= 3 \cdot 39

39

除以

339 = 13 \cdot 3

= 3 \cdot 3 \cdot 13

3, 13

都是质数，因此无法进一步因数分解

= 3 \cdot 3 \cdot 13

= 3^{2} \cdot 13

= 3^{2} \cdot 13

使用根式运算法则:

n ab = n a n b

= 13 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 313

u_{1, 2} = \frac{- ( - 11 ) \pm 3 13}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- ( - 11 ) + 3 13}{2 \cdot 1}, u_{2} = \frac{- ( - 11 ) - 3 13}{2 \cdot 1}

u = \frac{- ( - 11 ) + 3 13}{2 \cdot 1} : \frac{11 + 3 13}{2}

\frac{- ( - 11 ) + 3 13}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{11 + 3 13}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{11 + 3 13}{2}

u = \frac{- ( - 11 ) - 3 13}{2 \cdot 1} : \frac{11 - 3 13}{2}

\frac{- ( - 11 ) - 3 13}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{11 - 3 13}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{11 - 3 13}{2}

二次方程组的解是:

u = \frac{11 + 3 13}{2}, u = \frac{11 - 3 13}{2}

u = \frac{11 + 3 13}{2}, u = \frac{11 - 3 13}{2}

验证解

找到无定义的点（奇点）:

u = 0

取

- 11 + \frac{1}{u} + 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 = \frac{11 + 3 13}{2}, u = \frac{11 - 3 13}{2}

u = sec (2 x)

代回

sec (2 x) = \frac{11 + 3 13}{2}, sec (2 x) = \frac{11 - 3 13}{2}

sec (2 x) = \frac{11 + 3 13}{2}, sec (2 x) = \frac{11 - 3 13}{2}

sec (2 x) = \frac{11 + 3 13}{2} : x = \frac{arcsec ( \frac{11 + 3 13}{2} )}{2} + πn, x = π - \frac{arcsec ( \frac{11 + 3 13}{2} )}{2} + πn

sec (2 x) = \frac{11 + 3 13}{2}

使用反三角函数性质

sec (2 x) = \frac{11 + 3 13}{2}

sec (2 x) = \frac{11 + 3 13}{2}

的通解

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

2 x = arcsec (\frac{11 + 3 13}{2}) + 2 πn, 2 x = 2 π - arcsec (\frac{11 + 3 13}{2}) + 2 πn

2 x = arcsec (\frac{11 + 3 13}{2}) + 2 πn, 2 x = 2 π - arcsec (\frac{11 + 3 13}{2}) + 2 πn

解

2 x = arcsec (\frac{11 + 3 13}{2}) + 2 πn : x = \frac{arcsec ( \frac{11 + 3 13}{2} )}{2} + πn

2 x = arcsec (\frac{11 + 3 13}{2}) + 2 πn

两边除以

2

2 x = arcsec (\frac{11 + 3 13}{2}) + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{arcsec ( \frac{11 + 3 13}{2} )}{2} + \frac{2 πn}{2}

化简

x = \frac{arcsec ( \frac{11 + 3 13}{2} )}{2} + πn

x = \frac{arcsec ( \frac{11 + 3 13}{2} )}{2} + πn

解

2 x = 2 π - arcsec (\frac{11 + 3 13}{2}) + 2 πn : x = π - \frac{arcsec ( \frac{11 + 3 13}{2} )}{2} + πn

2 x = 2 π - arcsec (\frac{11 + 3 13}{2}) + 2 πn

两边除以

2

2 x = 2 π - arcsec (\frac{11 + 3 13}{2}) + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{2 π}{2} - \frac{arcsec ( \frac{11 + 3 13}{2} )}{2} + \frac{2 πn}{2}

化简

x = π - \frac{arcsec ( \frac{11 + 3 13}{2} )}{2} + πn

x = π - \frac{arcsec ( \frac{11 + 3 13}{2} )}{2} + πn

x = \frac{arcsec ( \frac{11 + 3 13}{2} )}{2} + πn, x = π - \frac{arcsec ( \frac{11 + 3 13}{2} )}{2} + πn

sec (2 x) = \frac{11 - 3 13}{2} :

无解

sec (2 x) = \frac{11 - 3 13}{2}

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

无解

合并所有解

x = \frac{arcsec ( \frac{11 + 3 13}{2} )}{2} + πn, x = π - \frac{arcsec ( \frac{11 + 3 13}{2} )}{2} + πn

以小数形式表示解

x = \frac{1.47899 \dots}{2} + πn, x = π - \frac{1.47899 \dots}{2} + πn

作图

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