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受欢迎的三角函数 >

3cos(2x)-7cos(x)+5=0

解答

3 cos (2 x) - 7 cos (x) + 5 = 0

解答

x = 0.84106 \dots + 2 πn, x = 2 π - 0.84106 \dots + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

+1

度数

x = 48.18968 \dots^{\circ} + 36 0^{\circ} n, x = 311.81031 \dots^{\circ} + 36 0^{\circ} n, x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

求解步骤

3 cos (2 x) - 7 cos (x) + 5 = 0

使用三角恒等式改写

5 + 3 cos (2 x) - 7 cos (x)

使用倍角公式:

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

= 5 + 3 (2 cos^{2} (x) - 1) - 7 cos (x)

化简

5 + 3 (2 cos^{2} (x) - 1) - 7 cos (x) : 6 cos^{2} (x) - 7 cos (x) + 2

5 + 3 (2 cos^{2} (x) - 1) - 7 cos (x)

乘开

3 (2 cos^{2} (x) - 1) : 6 cos^{2} (x) - 3

3 (2 cos^{2} (x) - 1)

使用分配律:

a (b - c) = ab - a c

a = 3, b = 2 cos^{2} (x), c = 1

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

化简

3 \cdot 2 cos^{2} (x) - 3 \cdot 1 : 6 cos^{2} (x) - 3

3 \cdot 2 cos^{2} (x) - 3 \cdot 1

数字相乘:

3 \cdot 2 = 6

= 6 cos^{2} (x) - 3 \cdot 1

数字相乘:

3 \cdot 1 = 3

= 6 cos^{2} (x) - 3

= 6 cos^{2} (x) - 3

= 5 + 6 cos^{2} (x) - 3 - 7 cos (x)

化简

5 + 6 cos^{2} (x) - 3 - 7 cos (x) : 6 cos^{2} (x) - 7 cos (x) + 2

5 + 6 cos^{2} (x) - 3 - 7 cos (x)

对同类项分组

= 6 cos^{2} (x) - 7 cos (x) + 5 - 3

数字相加/相减:

5 - 3 = 2

= 6 cos^{2} (x) - 7 cos (x) + 2

= 6 cos^{2} (x) - 7 cos (x) + 2

= 6 cos^{2} (x) - 7 cos (x) + 2

2 + 6 cos^{2} (x) - 7 cos (x) = 0

用替代法求解

2 + 6 cos^{2} (x) - 7 cos (x) = 0

令

: cos (x) = u

2 + 6 u^{2} - 7 u = 0

2 + 6 u^{2} - 7 u = 0 : u = \frac{2}{3}, u = \frac{1}{2}

2 + 6 u^{2} - 7 u = 0

改写成标准形式

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

6 u^{2} - 7 u + 2 = 0

使用求根公式求解

6 u^{2} - 7 u + 2 = 0

二次方程求根公式:

若

a = 6, b = - 7, c = 2

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 6 \cdot 2}{2 \cdot 6}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 6 \cdot 2}{2 \cdot 6}

(- 7)^{2} - 4 \cdot 6 \cdot 2 = 1

(- 7)^{2} - 4 \cdot 6 \cdot 2

使用指数法则:

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

若

n

是偶数

(- 7)^{2} = 7^{2}

= 7^{2} - 4 \cdot 6 \cdot 2

数字相乘:

4 \cdot 6 \cdot 2 = 48

= 7^{2} - 48

7^{2} = 49

= 49 - 48

数字相减:

49 - 48 = 1

= 1

使用法则

1 = 1

= 1

u_{1, 2} = \frac{- ( - 7 ) \pm 1}{2 \cdot 6}

将解分隔开

u_{1} = \frac{- ( - 7 ) + 1}{2 \cdot 6}, u_{2} = \frac{- ( - 7 ) - 1}{2 \cdot 6}

u = \frac{- ( - 7 ) + 1}{2 \cdot 6} : \frac{2}{3}

\frac{- ( - 7 ) + 1}{2 \cdot 6}

使用法则

- (- a) = a

= \frac{7 + 1}{2 \cdot 6}

数字相加:

7 + 1 = 8

= \frac{8}{2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{8}{12}

约分:

4

= \frac{2}{3}

u = \frac{- ( - 7 ) - 1}{2 \cdot 6} : \frac{1}{2}

\frac{- ( - 7 ) - 1}{2 \cdot 6}

使用法则

- (- a) = a

= \frac{7 - 1}{2 \cdot 6}

数字相减:

7 - 1 = 6

= \frac{6}{2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{6}{12}

约分:

6

= \frac{1}{2}

二次方程组的解是:

u = \frac{2}{3}, u = \frac{1}{2}

u = cos (x)

代回

cos (x) = \frac{2}{3}, cos (x) = \frac{1}{2}

cos (x) = \frac{2}{3}, cos (x) = \frac{1}{2}

cos (x) = \frac{2}{3} : x = arccos (\frac{2}{3}) + 2 πn, x = 2 π - arccos (\frac{2}{3}) + 2 πn

cos (x) = \frac{2}{3}

使用反三角函数性质

cos (x) = \frac{2}{3}

cos (x) = \frac{2}{3}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{2}{3}) + 2 πn, x = 2 π - arccos (\frac{2}{3}) + 2 πn

x = arccos (\frac{2}{3}) + 2 πn, x = 2 π - arccos (\frac{2}{3}) + 2 πn

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = \frac{1}{2}

cos (x) = \frac{1}{2}

的通解

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} 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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

合并所有解

x = arccos (\frac{2}{3}) + 2 πn, x = 2 π - arccos (\frac{2}{3}) + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

以小数形式表示解

x = 0.84106 \dots + 2 πn, x = 2 π - 0.84106 \dots + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

作图

流行的例子

6 cos (x) + 3 = 0

sin (x) = \frac{7}{12}

sin (x) = \frac{7}{13}

cos(x)+2cos(x+240)=1

cos (x) + 2 cos (x + 24 0^{\circ}) = 1

cos (x π) = 0