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受欢迎的三角函数 >

2cos(θ)-1=sec(θ)

解答

2 cos (θ) - 1 = sec (θ)

解答

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn, θ = 2 πn

+1

度数

θ = 12 0^{\circ} + 36 0^{\circ} n, θ = 24 0^{\circ} + 36 0^{\circ} n, θ = 0^{\circ} + 36 0^{\circ} n

求解步骤

2 cos (θ) - 1 = sec (θ)

两边减去

sec (θ)

2 cos (θ) - 1 - sec (θ) = 0

使用三角恒等式改写

- 1 - sec (θ) + 2 cos (θ)

使用基本三角恒等式:

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

= - 1 - sec (θ) + 2 \cdot \frac{1}{sec ( θ )}

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

2 \cdot \frac{1}{sec ( θ )}

分式相乘:

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

= \frac{1 \cdot 2}{sec ( θ )}

数字相乘:

1 \cdot 2 = 2

= \frac{2}{sec ( θ )}

= - 1 - sec (θ) + \frac{2}{sec ( θ )}

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

用替代法求解

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

令

: sec (θ) = u

- 1 + \frac{2}{u} - u = 0

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

- 1 + \frac{2}{u} - u = 0

在两边乘以

u

- 1 + \frac{2}{u} - u = 0

在两边乘以

u

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

化简

- 1 \cdot u + \frac{2}{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{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

- 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 + 2 - u^{2} = 0

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

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

解

- u + 2 - u^{2} = 0 : u = - 2, u = 1

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = - 1, b = - 1, c = 2

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

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

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

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

使用法则

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

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

若

n

是偶数

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

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 1 \cdot 2 = 8

4 \cdot 1 \cdot 2

数字相乘:

4 \cdot 1 \cdot 2 = 8

= 8

= 1 + 8

数字相加:

1 + 8 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 3

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

将解分隔开

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

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

\frac{- ( - 1 ) + 3}{2 ( - 1 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{1 + 3}{- 2 \cdot 1}

数字相加:

1 + 3 = 4

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

数字相乘:

2 \cdot 1 = 2

= \frac{4}{- 2}

使用分式法则:

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

= - \frac{4}{2}

数字相除:

\frac{4}{2} = 2

= - 2

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

\frac{- ( - 1 ) - 3}{2 ( - 1 )}

去除括号:

(- a) = - a, - (- a) = a

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

数字相减:

1 - 3 = - 2

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

数字相乘:

2 \cdot 1 = 2

= \frac{- 2}{- 2}

使用分式法则:

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

= \frac{2}{2}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

u = - 2, u = 1

u = - 2, u = 1

验证解

找到无定义的点（奇点）:

u = 0

取

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

u = sec (θ)

代回

sec (θ) = - 2, sec (θ) = 1

sec (θ) = - 2, sec (θ) = 1

sec (θ) = - 2 : θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

sec (θ) = - 2

sec (θ) = - 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}

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

sec (θ) = 1 : θ = 2 πn

sec (θ) = 1

sec (θ) = 1

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}

θ = 0 + 2 πn

θ = 0 + 2 πn

解

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

合并所有解

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn, θ = 2 πn

作图

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