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受欢迎的三角函数 >

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

解答

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

解答

x = 0.93592 \dots + 2 πn, x = 2 π - 0.93592 \dots + 2 πn, x = 2.57376 \dots + 2 πn, x = - 2.57376 \dots + 2 πn

+1

度数

x = 53.62480 \dots^{\circ} + 36 0^{\circ} n, x = 306.37519 \dots^{\circ} + 36 0^{\circ} n, x = 147.46577 \dots^{\circ} + 36 0^{\circ} n, x = - 147.46577 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

使用三角恒等式改写

cos (x) + 2 cos (2 x)

使用倍角公式:

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

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

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

用替代法求解

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

令

: cos (x) = u

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

u + (- 1 + 2 u^{2}) \cdot 2 = 0 : u = \frac{- 1 + 33}{8}, u = \frac{- 1 - 33}{8}

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

展开

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

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

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

乘开

2 (- 1 + 2 u^{2}) : - 2 + 4 u^{2}

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

使用分配律:

a (b + c) = ab + a c

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

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

使用加减运算法则

+ (- a) = - a

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

化简

- 2 \cdot 1 + 2 \cdot 2 u^{2} : - 2 + 4 u^{2}

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

数字相乘:

2 \cdot 1 = 2

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

数字相乘:

2 \cdot 2 = 4

= - 2 + 4 u^{2}

= - 2 + 4 u^{2}

= u - 2 + 4 u^{2}

u - 2 + 4 u^{2} = 0

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = 4, b = 1, c = - 2

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

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

1^{2} - 4 \cdot 4 (- 2) = 33

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 4 (- 2)

使用法则

- (- a) = a

= 1 + 4 \cdot 4 \cdot 2

数字相乘:

4 \cdot 4 \cdot 2 = 32

= 1 + 32

数字相加:

1 + 32 = 33

= 33

u_{1, 2} = \frac{- 1 \pm 33}{2 \cdot 4}

将解分隔开

u_{1} = \frac{- 1 + 33}{2 \cdot 4}, u_{2} = \frac{- 1 - 33}{2 \cdot 4}

u = \frac{- 1 + 33}{2 \cdot 4} : \frac{- 1 + 33}{8}

\frac{- 1 + 33}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 1 + 33}{8}

u = \frac{- 1 - 33}{2 \cdot 4} : \frac{- 1 - 33}{8}

\frac{- 1 - 33}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 1 - 33}{8}

二次方程组的解是:

u = \frac{- 1 + 33}{8}, u = \frac{- 1 - 33}{8}

u = cos (x)

代回

cos (x) = \frac{- 1 + 33}{8}, cos (x) = \frac{- 1 - 33}{8}

cos (x) = \frac{- 1 + 33}{8}, cos (x) = \frac{- 1 - 33}{8}

cos (x) = \frac{- 1 + 33}{8} : x = arccos (\frac{- 1 + 33}{8}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 33}{8}) + 2 πn

cos (x) = \frac{- 1 + 33}{8}

使用反三角函数性质

cos (x) = \frac{- 1 + 33}{8}

cos (x) = \frac{- 1 + 33}{8}

的通解

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

x = arccos (\frac{- 1 + 33}{8}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 33}{8}) + 2 πn

x = arccos (\frac{- 1 + 33}{8}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 33}{8}) + 2 πn

cos (x) = \frac{- 1 - 33}{8} : x = arccos (\frac{- 1 - 33}{8}) + 2 πn, x = - arccos (\frac{- 1 - 33}{8}) + 2 πn

cos (x) = \frac{- 1 - 33}{8}

使用反三角函数性质

cos (x) = \frac{- 1 - 33}{8}

cos (x) = \frac{- 1 - 33}{8}

的通解

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

x = arccos (\frac{- 1 - 33}{8}) + 2 πn, x = - arccos (\frac{- 1 - 33}{8}) + 2 πn

x = arccos (\frac{- 1 - 33}{8}) + 2 πn, x = - arccos (\frac{- 1 - 33}{8}) + 2 πn

合并所有解

x = arccos (\frac{- 1 + 33}{8}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 33}{8}) + 2 πn, x = arccos (\frac{- 1 - 33}{8}) + 2 πn, x = - arccos (\frac{- 1 - 33}{8}) + 2 πn

以小数形式表示解

x = 0.93592 \dots + 2 πn, x = 2 π - 0.93592 \dots + 2 πn, x = 2.57376 \dots + 2 πn, x = - 2.57376 \dots + 2 πn

作图

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