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受欢迎的三角函数 >

3cos(2x)+cos(x)=0

解答

3 cos (2 x) + cos (x) = 0

解答

x = 0.89095 \dots + 2 πn, x = 2 π - 0.89095 \dots + 2 πn, x = 2.49035 \dots + 2 πn, x = - 2.49035 \dots + 2 πn

+1

度数

x = 51.04815 \dots^{\circ} + 36 0^{\circ} n, x = 308.95184 \dots^{\circ} + 36 0^{\circ} n, x = 142.68678 \dots^{\circ} + 36 0^{\circ} n, x = - 142.68678 \dots^{\circ} + 36 0^{\circ} n

求解步骤

3 cos (2 x) + cos (x) = 0

使用三角恒等式改写

cos (x) + 3 cos (2 x)

使用倍角公式:

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

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

cos (x) + (- 1 + 2 cos^{2} (x)) \cdot 3 = 0

用替代法求解

cos (x) + (- 1 + 2 cos^{2} (x)) \cdot 3 = 0

令

: cos (x) = u

u + (- 1 + 2 u^{2}) \cdot 3 = 0

u + (- 1 + 2 u^{2}) \cdot 3 = 0 : u = \frac{- 1 + 73}{12}, u = \frac{- 1 - 73}{12}

u + (- 1 + 2 u^{2}) \cdot 3 = 0

展开

u + (- 1 + 2 u^{2}) \cdot 3 : u - 3 + 6 u^{2}

u + (- 1 + 2 u^{2}) \cdot 3

= u + 3 (- 1 + 2 u^{2})

乘开

3 (- 1 + 2 u^{2}) : - 3 + 6 u^{2}

3 (- 1 + 2 u^{2})

使用分配律:

a (b + c) = ab + a c

a = 3, b = - 1, c = 2 u^{2}

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

使用加减运算法则

+ (- a) = - a

= - 3 \cdot 1 + 3 \cdot 2 u^{2}

化简

- 3 \cdot 1 + 3 \cdot 2 u^{2} : - 3 + 6 u^{2}

- 3 \cdot 1 + 3 \cdot 2 u^{2}

数字相乘:

3 \cdot 1 = 3

= - 3 + 3 \cdot 2 u^{2}

数字相乘:

3 \cdot 2 = 6

= - 3 + 6 u^{2}

= - 3 + 6 u^{2}

= u - 3 + 6 u^{2}

u - 3 + 6 u^{2} = 0

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = 6, b = 1, c = - 3

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 6 ( - 3 )}{2 \cdot 6}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 6 ( - 3 )}{2 \cdot 6}

1^{2} - 4 \cdot 6 (- 3) = 73

1^{2} - 4 \cdot 6 (- 3)

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 6 (- 3)

使用法则

- (- a) = a

= 1 + 4 \cdot 6 \cdot 3

数字相乘:

4 \cdot 6 \cdot 3 = 72

= 1 + 72

数字相加:

1 + 72 = 73

= 73

u_{1, 2} = \frac{- 1 \pm 73}{2 \cdot 6}

将解分隔开

u_{1} = \frac{- 1 + 73}{2 \cdot 6}, u_{2} = \frac{- 1 - 73}{2 \cdot 6}

u = \frac{- 1 + 73}{2 \cdot 6} : \frac{- 1 + 73}{12}

\frac{- 1 + 73}{2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{- 1 + 73}{12}

u = \frac{- 1 - 73}{2 \cdot 6} : \frac{- 1 - 73}{12}

\frac{- 1 - 73}{2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{- 1 - 73}{12}

二次方程组的解是:

u = \frac{- 1 + 73}{12}, u = \frac{- 1 - 73}{12}

u = cos (x)

代回

cos (x) = \frac{- 1 + 73}{12}, cos (x) = \frac{- 1 - 73}{12}

cos (x) = \frac{- 1 + 73}{12}, cos (x) = \frac{- 1 - 73}{12}

cos (x) = \frac{- 1 + 73}{12} : x = arccos (\frac{- 1 + 73}{12}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 73}{12}) + 2 πn

cos (x) = \frac{- 1 + 73}{12}

使用反三角函数性质

cos (x) = \frac{- 1 + 73}{12}

cos (x) = \frac{- 1 + 73}{12}

的通解

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

x = arccos (\frac{- 1 + 73}{12}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 73}{12}) + 2 πn

x = arccos (\frac{- 1 + 73}{12}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 73}{12}) + 2 πn

cos (x) = \frac{- 1 - 73}{12} : x = arccos (\frac{- 1 - 73}{12}) + 2 πn, x = - arccos (\frac{- 1 - 73}{12}) + 2 πn

cos (x) = \frac{- 1 - 73}{12}

使用反三角函数性质

cos (x) = \frac{- 1 - 73}{12}

cos (x) = \frac{- 1 - 73}{12}

的通解

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

x = arccos (\frac{- 1 - 73}{12}) + 2 πn, x = - arccos (\frac{- 1 - 73}{12}) + 2 πn

x = arccos (\frac{- 1 - 73}{12}) + 2 πn, x = - arccos (\frac{- 1 - 73}{12}) + 2 πn

合并所有解

x = arccos (\frac{- 1 + 73}{12}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 73}{12}) + 2 πn, x = arccos (\frac{- 1 - 73}{12}) + 2 πn, x = - arccos (\frac{- 1 - 73}{12}) + 2 πn

以小数形式表示解

x = 0.89095 \dots + 2 πn, x = 2 π - 0.89095 \dots + 2 πn, x = 2.49035 \dots + 2 πn, x = - 2.49035 \dots + 2 πn

作图

流行的例子

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