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受欢迎的三角函数 >

cos(3x)-21cos(x)+16=0

解答

cos (3 x) - 21 cos (x) + 16 = 0

解答

x = 0.74946 \dots + 2 πn, x = 2 π - 0.74946 \dots + 2 πn

+1

度数

x = 42.94140 \dots^{\circ} + 36 0^{\circ} n, x = 317.05859 \dots^{\circ} + 36 0^{\circ} n

求解步骤

cos (3 x) - 21 cos (x) + 16 = 0

使用三角恒等式改写

16 + cos (3 x) - 21 cos (x)

cos (3 x) = 4 cos^{3} (x) - 3 cos (x)

cos (3 x)

使用三角恒等式改写

cos (3 x)

改写为

= cos (2 x + x)

使用角和恒等式:

cos (s + t) = cos (s) cos (t) - sin (s) sin (t)

= cos (2 x) cos (x) - sin (2 x) sin (x)

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

= cos (2 x) cos (x) - 2 sin (x) cos (x) sin (x)

化简

cos (2 x) cos (x) - 2 sin (x) cos (x) sin (x) : cos (x) cos (2 x) - 2 sin^{2} (x) cos (x)

cos (2 x) cos (x) - 2 sin (x) cos (x) sin (x)

2 sin (x) cos (x) sin (x) = 2 sin^{2} (x) cos (x)

2 sin (x) cos (x) sin (x)

使用指数法则:

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

sin (x) sin (x) = sin^{1 + 1} (x)

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

数字相加:

1 + 1 = 2

= 2 cos (x) sin^{2} (x)

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

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

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

使用倍角公式:

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

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

使用毕达哥拉斯恒等式:

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

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

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

乘开

(2 cos^{2} (x) - 1) cos (x) - 2 (1 - cos^{2} (x)) cos (x) : 4 cos^{3} (x) - 3 cos (x)

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

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

乘开

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

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

使用分配律:

a (b - c) = ab - a c

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

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

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

化简

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

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

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

2 cos^{2} (x) cos (x)

使用指数法则:

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

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

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

数字相加:

2 + 1 = 3

= 2 cos^{3} (x)

1 \cdot cos (x) = cos (x)

1 cos (x)

乘以:

1 \cdot cos (x) = cos (x)

= cos (x)

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

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

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

乘开

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

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

使用分配律:

a (b - c) = ab - a c

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

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

使用加减运算法则

- (- a) = a

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

化简

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

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

2 \cdot 1 \cdot cos (x) = 2 cos (x)

2 \cdot 1 cos (x)

数字相乘:

2 \cdot 1 = 2

= 2 cos (x)

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

2 cos^{2} (x) cos (x)

使用指数法则:

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

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

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

数字相加:

2 + 1 = 3

= 2 cos^{3} (x)

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

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

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

化简

2 cos^{3} (x) - cos (x) - 2 cos (x) + 2 cos^{3} (x) : 4 cos^{3} (x) - 3 cos (x)

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

对同类项分组

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

同类项相加:

2 cos^{3} (x) + 2 cos^{3} (x) = 4 cos^{3} (x)

= 4 cos^{3} (x) - cos (x) - 2 cos (x)

同类项相加:

- cos (x) - 2 cos (x) = - 3 cos (x)

= 4 cos^{3} (x) - 3 cos (x)

= 4 cos^{3} (x) - 3 cos (x)

= 4 cos^{3} (x) - 3 cos (x)

= 16 + 4 cos^{3} (x) - 3 cos (x) - 21 cos (x)

化简

= 16 + 4 cos^{3} (x) - 24 cos (x)

16 - 24 cos (x) + 4 cos^{3} (x) = 0

用替代法求解

16 - 24 cos (x) + 4 cos^{3} (x) = 0

令

: cos (x) = u

16 - 24 u + 4 u^{3} = 0

16 - 24 u + 4 u^{3} = 0 : u = 2, u = - 1 + 3, u = - 1 - 3

16 - 24 u + 4 u^{3} = 0

改写成标准形式

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

4 u^{3} - 24 u + 16 = 0

因式分解

4 u^{3} - 24 u + 16 : 4 (u - 2) (u^{2} + 2 u - 2)

4 u^{3} - 24 u + 16

因式分解出通项

4 : 4 (u^{3} - 6 u + 4)

4 u^{3} - 24 u + 16

将

16

改写为

4 \cdot 4

将

24

改写为

4 \cdot 6

= 4 u^{3} - 4 \cdot 6 u + 4 \cdot 4

因式分解出通项

4

= 4 (u^{3} - 6 u + 4)

= 4 (u^{3} - 6 u + 4)

分解

u^{3} - 6 u + 4 : (u - 2) (u^{2} + 2 u - 2)

u^{3} - 6 u + 4

使用有理根定理

a_{0} = 4, a_{n} = 1

a_{0}

的除数

: 1, 2, 4, a_{n}

的除数

: 1

因此，检验以下有理数:

\pm \frac{1 , 2 , 4}{1}

\frac{2}{1}

是表达式的根，所以因式分解

u - 2

= (u - 2) \frac{u ^{3} - 6 u + 4}{u - 2}

\frac{u ^{3} - 6 u + 4}{u - 2} = u^{2} + 2 u - 2

\frac{u ^{3} - 6 u + 4}{u - 2}

对

\frac{u ^{3} - 6 u + 4}{u - 2}

做除法:

\frac{u ^{3} - 6 u + 4}{u - 2} = u^{2} + \frac{2 u ^{2} - 6 u + 4}{u - 2}

将分子

u^{3} - 6 u + 4

与除数

u - 2

的首项系数相除：

\frac{u ^{3}}{u} = u^{2}

商 = u^{2}

将

u - 2

乘以

u^{2} : u^{3} - 2 u^{2}

将

u^{3} - 6 u + 4

减去

u^{3} - 2 u^{2}

得到新的余数

余数 = 2 u^{2} - 6 u + 4

因此

\frac{u ^{3} - 6 u + 4}{u - 2} = u^{2} + \frac{2 u ^{2} - 6 u + 4}{u - 2}

= u^{2} + \frac{2 u ^{2} - 6 u + 4}{u - 2}

对

\frac{2 u ^{2} - 6 u + 4}{u - 2}

做除法:

\frac{2 u ^{2} - 6 u + 4}{u - 2} = 2 u + \frac{- 2 u + 4}{u - 2}

将分子

2 u^{2} - 6 u + 4

与除数

u - 2

的首项系数相除：

\frac{2 u ^{2}}{u} = 2 u

商 = 2 u

将

u - 2

乘以

2 u : 2 u^{2} - 4 u

将

2 u^{2} - 6 u + 4

减去

2 u^{2} - 4 u

得到新的余数

余数 = - 2 u + 4

因此

\frac{2 u ^{2} - 6 u + 4}{u - 2} = 2 u + \frac{- 2 u + 4}{u - 2}

= u^{2} + 2 u + \frac{- 2 u + 4}{u - 2}

对

\frac{- 2 u + 4}{u - 2}

做除法:

\frac{- 2 u + 4}{u - 2} = - 2

将分子

- 2 u + 4

与除数

u - 2

的首项系数相除：

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

商 = - 2

将

u - 2

乘以

- 2 : - 2 u + 4

将

- 2 u + 4

减去

- 2 u + 4

得到新的余数

余数 = 0

因此

\frac{- 2 u + 4}{u - 2} = - 2

= u^{2} + 2 u - 2

= u^{2} + 2 u - 2

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

= 4 (u - 2) (u^{2} + 2 u - 2)

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

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

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

解

u - 2 = 0 : u = 2

u - 2 = 0

将

2

到右边

u - 2 = 0

两边加上

2

u - 2 + 2 = 0 + 2

化简

u = 2

u = 2

解

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

2^{2} - 4 \cdot 1 \cdot (- 2) = 23

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

使用法则

- (- a) = a

= 2^{2} + 4 \cdot 1 \cdot 2

数字相乘:

4 \cdot 1 \cdot 2 = 8

= 2^{2} + 8

2^{2} = 4

= 4 + 8

数字相加:

4 + 8 = 12

= 12

12

质因数分解:

2^{2} \cdot 3

12

12

除以

212 = 6 \cdot 2

= 2 \cdot 6

6

除以

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 3

= 2^{2} \cdot 3

= 2^{2} \cdot 3

使用根式运算法则:

n ab = n a n b

= 3 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 23

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

将解分隔开

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

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

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

数字相乘:

2 \cdot 1 = 2

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

分解

- 2 + 23 : 2 (- 1 + 3)

- 2 + 23

改写为

= - 2 \cdot 1 + 23

因式分解出通项

2

= 2 (- 1 + 3)

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

数字相除:

\frac{2}{2} = 1

= - 1 + 3

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

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

数字相乘:

2 \cdot 1 = 2

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

分解

- 2 - 23 : - 2 (1 + 3)

- 2 - 23

改写为

= - 2 \cdot 1 - 23

因式分解出通项

2

= - 2 (1 + 3)

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

数字相除:

\frac{2}{2} = 1

= - (1 + 3)

取负

- (1 + 3) = - 1 - 3

= - 1 - 3

二次方程组的解是:

u = - 1 + 3, u = - 1 - 3

解为

u = 2, u = - 1 + 3, u = - 1 - 3

u = cos (x)

代回

cos (x) = 2, cos (x) = - 1 + 3, cos (x) = - 1 - 3

cos (x) = 2, cos (x) = - 1 + 3, cos (x) = - 1 - 3

cos (x) = 2 :

无解

cos (x) = 2

- 1 \leq cos (x) \leq 1

无解

cos (x) = - 1 + 3 : x = arccos (- 1 + 3) + 2 πn, x = 2 π - arccos (- 1 + 3) + 2 πn

cos (x) = - 1 + 3

使用反三角函数性质

cos (x) = - 1 + 3

cos (x) = - 1 + 3

的通解

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

x = arccos (- 1 + 3) + 2 πn, x = 2 π - arccos (- 1 + 3) + 2 πn

x = arccos (- 1 + 3) + 2 πn, x = 2 π - arccos (- 1 + 3) + 2 πn

cos (x) = - 1 - 3 :

无解

cos (x) = - 1 - 3

- 1 \leq cos (x) \leq 1

无解

合并所有解

x = arccos (- 1 + 3) + 2 πn, x = 2 π - arccos (- 1 + 3) + 2 πn

以小数形式表示解

x = 0.74946 \dots + 2 πn, x = 2 π - 0.74946 \dots + 2 πn

作图

流行的例子

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sin^{5} (x) - sin (x) = 0

tan(x)=(95.75)/4

tan (x) = \frac{95.75}{4}

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cos(x-15^0)=((sqrt(2)))/2

cos (x - 1 5^{0}) = \frac{( 2 )}{2}

sin(x)=1-cos^x(x)

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