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受欢迎的三角函数 >

2cos(θ)-2sec(θ)-3=0

解答

2 cos (θ) - 2 sec (θ) - 3 = 0

解答

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

+1

度数

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

求解步骤

2 cos (θ) - 2 sec (θ) - 3 = 0

使用三角恒等式改写

- 3 + 2 cos (θ) - 2 sec (θ)

使用基本三角恒等式:

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

= - 3 + 2 \cdot \frac{1}{sec ( θ )} - 2 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 )}

= - 3 + \frac{2}{sec ( θ )} - 2 sec (θ)

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

用替代法求解

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

令

: sec (θ) = u

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

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

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

在两边乘以

u

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

在两边乘以

u

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

化简

- 3 u + \frac{2}{u} u - 2 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

- 2 uu : - 2 u^{2}

- 2 uu

使用指数法则:

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

uu = u^{1 + 1}

= - 2 u^{1 + 1}

数字相加:

1 + 1 = 2

= - 2 u^{2}

化简

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

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

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

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

解

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = - 2, b = - 3, c = 2

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

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

(- 3)^{2} - 4 (- 2) \cdot 2 = 5

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

使用法则

- (- a) = a

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

使用指数法则:

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

若

n

是偶数

(- 3)^{2} = 3^{2}

= 3^{2} + 4 \cdot 2 \cdot 2

数字相乘:

4 \cdot 2 \cdot 2 = 16

= 3^{2} + 16

3^{2} = 9

= 9 + 16

数字相加:

9 + 16 = 25

= 25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 5

u_{1, 2} = \frac{- ( - 3 ) \pm 5}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- ( - 3 ) + 5}{2 ( - 2 )}, u_{2} = \frac{- ( - 3 ) - 5}{2 ( - 2 )}

u = \frac{- ( - 3 ) + 5}{2 ( - 2 )} : - 2

\frac{- ( - 3 ) + 5}{2 ( - 2 )}

去除括号:

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

= \frac{3 + 5}{- 2 \cdot 2}

数字相加:

3 + 5 = 8

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

数字相乘:

2 \cdot 2 = 4

= \frac{8}{- 4}

使用分式法则:

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

= - \frac{8}{4}

数字相除:

\frac{8}{4} = 2

= - 2

u = \frac{- ( - 3 ) - 5}{2 ( - 2 )} : \frac{1}{2}

\frac{- ( - 3 ) - 5}{2 ( - 2 )}

去除括号:

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

= \frac{3 - 5}{- 2 \cdot 2}

数字相减:

3 - 5 = - 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

使用分式法则:

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

= \frac{2}{4}

约分:

2

= \frac{1}{2}

二次方程组的解是:

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

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

验证解

找到无定义的点（奇点）:

u = 0

取

- 3 + \frac{2}{u} - 2 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}{2}

u = sec (θ)

代回

sec (θ) = - 2, sec (θ) = \frac{1}{2}

sec (θ) = - 2, sec (θ) = \frac{1}{2}

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) = \frac{1}{2} :

无解

sec (θ) = \frac{1}{2}

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

无解

合并所有解

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

作图

流行的例子

2sin^2(x)-3cos(x)=0

2 sin^{2} (x) - 3 cos (x) = 0

sin (θ) = \frac{2}{9}

-2sin^2(θ)+9sin(θ)=8sin(θ)-1

- 2 sin^{2} (θ) + 9 sin (θ) = 8 sin (θ) - 1

sin^2(x)-sin(x)-1=0

sin^{2} (x) - sin (x) - 1 = 0

2 sec (x) - 4 = 0