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受欢迎的三角函数 >

3cos(x)+1=sec(x)

解答

3 cos (x) + 1 = sec (x)

解答

x = 2.44587 \dots + 2 πn, x = - 2.44587 \dots + 2 πn, x = 1.12158 \dots + 2 πn, x = 2 π - 1.12158 \dots + 2 πn

+1

度数

x = 140.13813 \dots^{\circ} + 36 0^{\circ} n, x = - 140.13813 \dots^{\circ} + 36 0^{\circ} n, x = 64.26187 \dots^{\circ} + 36 0^{\circ} n, x = 295.73812 \dots^{\circ} + 36 0^{\circ} n

求解步骤

3 cos (x) + 1 = sec (x)

两边减去

sec (x)

3 cos (x) + 1 - sec (x) = 0

使用三角恒等式改写

1 - sec (x) + 3 cos (x)

使用基本三角恒等式:

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

= 1 - sec (x) + 3 \cdot \frac{1}{sec ( x )}

3 \cdot \frac{1}{sec ( x )} = \frac{3}{sec ( x )}

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

分式相乘:

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

= \frac{1 \cdot 3}{sec ( x )}

数字相乘:

1 \cdot 3 = 3

= \frac{3}{sec ( x )}

= 1 - sec (x) + \frac{3}{sec ( x )}

1 + \frac{3}{sec ( x )} - sec (x) = 0

用替代法求解

1 + \frac{3}{sec ( x )} - sec (x) = 0

令

: sec (x) = u

1 + \frac{3}{u} - u = 0

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

1 + \frac{3}{u} - u = 0

在两边乘以

u

1 + \frac{3}{u} - u = 0

在两边乘以

u

1 \cdot u + \frac{3}{u} u - uu = 0 \cdot u

化简

1 \cdot u + \frac{3}{u} u - uu = 0 \cdot u

化简

1 \cdot u : u

1 \cdot u

乘以:

1 \cdot u = u

= u

化简

\frac{3}{u} u : 3

\frac{3}{u} u

分式相乘:

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

= \frac{3 u}{u}

约分:

u

= 3

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

u + 3 - u^{2} = 0

u + 3 - u^{2} = 0

u + 3 - u^{2} = 0

解

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

u + 3 - u^{2} = 0

改写成标准形式

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

- u^{2} + u + 3 = 0

使用求根公式求解

- u^{2} + u + 3 = 0

二次方程求根公式:

若

a = - 1, b = 1, c = 3

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) \cdot 3}{2 ( - 1 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) \cdot 3}{2 ( - 1 )}

1^{2} - 4 (- 1) \cdot 3 = 13

1^{2} - 4 (- 1) \cdot 3

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) \cdot 3

使用法则

- (- a) = a

= 1 + 4 \cdot 1 \cdot 3

数字相乘:

4 \cdot 1 \cdot 3 = 12

= 1 + 12

数字相加:

1 + 12 = 13

= 13

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

将解分隔开

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

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

\frac{- 1 + 13}{2 ( - 1 )}

去除括号:

(- a) = - a

= \frac{- 1 + 13}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 1 + 13}{- 2}

使用分式法则:

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

= - \frac{- 1 + 13}{2}

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

\frac{- 1 - 13}{2 ( - 1 )}

去除括号:

(- a) = - a

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

数字相乘:

2 \cdot 1 = 2

= \frac{- 1 - 13}{- 2}

使用分式法则:

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

- 1 - 13 = - (1 + 13)

= \frac{1 + 13}{2}

二次方程组的解是:

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

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

验证解

找到无定义的点（奇点）:

u = 0

取

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

u = sec (x)

代回

sec (x) = - \frac{- 1 + 13}{2}, sec (x) = \frac{1 + 13}{2}

sec (x) = - \frac{- 1 + 13}{2}, sec (x) = \frac{1 + 13}{2}

sec (x) = - \frac{- 1 + 13}{2} : x = arcsec (- \frac{- 1 + 13}{2}) + 2 πn, x = - arcsec (- \frac{- 1 + 13}{2}) + 2 πn

sec (x) = - \frac{- 1 + 13}{2}

使用反三角函数性质

sec (x) = - \frac{- 1 + 13}{2}

sec (x) = - \frac{- 1 + 13}{2}

的通解

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

x = arcsec (- \frac{- 1 + 13}{2}) + 2 πn, x = - arcsec (- \frac{- 1 + 13}{2}) + 2 πn

x = arcsec (- \frac{- 1 + 13}{2}) + 2 πn, x = - arcsec (- \frac{- 1 + 13}{2}) + 2 πn

sec (x) = \frac{1 + 13}{2} : x = arcsec (\frac{1 + 13}{2}) + 2 πn, x = 2 π - arcsec (\frac{1 + 13}{2}) + 2 πn

sec (x) = \frac{1 + 13}{2}

使用反三角函数性质

sec (x) = \frac{1 + 13}{2}

sec (x) = \frac{1 + 13}{2}

的通解

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

x = arcsec (\frac{1 + 13}{2}) + 2 πn, x = 2 π - arcsec (\frac{1 + 13}{2}) + 2 πn

x = arcsec (\frac{1 + 13}{2}) + 2 πn, x = 2 π - arcsec (\frac{1 + 13}{2}) + 2 πn

合并所有解

x = arcsec (- \frac{- 1 + 13}{2}) + 2 πn, x = - arcsec (- \frac{- 1 + 13}{2}) + 2 πn, x = arcsec (\frac{1 + 13}{2}) + 2 πn, x = 2 π - arcsec (\frac{1 + 13}{2}) + 2 πn

以小数形式表示解

x = 2.44587 \dots + 2 πn, x = - 2.44587 \dots + 2 πn, x = 1.12158 \dots + 2 πn, x = 2 π - 1.12158 \dots + 2 πn

作图

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