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受欢迎的三角函数 >

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

解答

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

解答

x = 2 πn, x = 1.71777 \dots + 2 πn, x = - 1.71777 \dots + 2 πn, x = 2.59356 \dots + 2 πn, x = - 2.59356 \dots + 2 πn

+1

度数

x = 0^{\circ} + 36 0^{\circ} n, x = 98.42105 \dots^{\circ} + 36 0^{\circ} n, x = - 98.42105 \dots^{\circ} + 36 0^{\circ} n, x = 148.60028 \dots^{\circ} + 36 0^{\circ} n, x = - 148.60028 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

1 + cos (x)

2 cos (3 x) - 1 - cos (x) = 0

使用三角恒等式改写

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

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

化简

- 1 - cos (x) + 2 (4 cos^{3} (x) - 3 cos (x)) : - 7 cos (x) + 8 cos^{3} (x) - 1

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

乘开

2 (4 cos^{3} (x) - 3 cos (x)) : 8 cos^{3} (x) - 6 cos (x)

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

使用分配律:

a (b - c) = ab - a c

a = 2, b = 4 cos^{3} (x), c = 3 cos (x)

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

化简

2 \cdot 4 cos^{3} (x) - 2 \cdot 3 cos (x) : 8 cos^{3} (x) - 6 cos (x)

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

数字相乘:

2 \cdot 4 = 8

= 8 cos^{3} (x) - 2 \cdot 3 cos (x)

数字相乘:

2 \cdot 3 = 6

= 8 cos^{3} (x) - 6 cos (x)

= 8 cos^{3} (x) - 6 cos (x)

= - 1 - cos (x) + 8 cos^{3} (x) - 6 cos (x)

化简

- 1 - cos (x) + 8 cos^{3} (x) - 6 cos (x) : - 7 cos (x) + 8 cos^{3} (x) - 1

- 1 - cos (x) + 8 cos^{3} (x) - 6 cos (x)

对同类项分组

= - cos (x) + 8 cos^{3} (x) - 6 cos (x) - 1

同类项相加:

- cos (x) - 6 cos (x) = - 7 cos (x)

= - 7 cos (x) + 8 cos^{3} (x) - 1

= - 7 cos (x) + 8 cos^{3} (x) - 1

= - 7 cos (x) + 8 cos^{3} (x) - 1

- 1 - 7 cos (x) + 8 cos^{3} (x) = 0

用替代法求解

- 1 - 7 cos (x) + 8 cos^{3} (x) = 0

令

: cos (x) = u

- 1 - 7 u + 8 u^{3} = 0

- 1 - 7 u + 8 u^{3} = 0 : u = 1, u = \frac{- 2 + 2}{4}, u = - \frac{2 + 2}{4}

- 1 - 7 u + 8 u^{3} = 0

改写成标准形式

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

8 u^{3} - 7 u - 1 = 0

因式分解

8 u^{3} - 7 u - 1 : (u - 1) (8 u^{2} + 8 u + 1)

8 u^{3} - 7 u - 1

使用有理根定理

a_{0} = 1, a_{n} = 8

a_{0}

的除数

: 1, a_{n}

的除数

: 1, 2, 4, 8

因此，检验以下有理数:

\pm \frac{1}{1 , 2 , 4 , 8}

\frac{1}{1}

是表达式的根，所以因式分解

u - 1

= (u - 1) \frac{8 u ^{3} - 7 u - 1}{u - 1}

\frac{8 u ^{3} - 7 u - 1}{u - 1} = 8 u^{2} + 8 u + 1

\frac{8 u ^{3} - 7 u - 1}{u - 1}

对

\frac{8 u ^{3} - 7 u - 1}{u - 1}

做除法:

\frac{8 u ^{3} - 7 u - 1}{u - 1} = 8 u^{2} + \frac{8 u ^{2} - 7 u - 1}{u - 1}

将分子

8 u^{3} - 7 u - 1

与除数

u - 1

的首项系数相除：

\frac{8 u ^{3}}{u} = 8 u^{2}

商 = 8 u^{2}

将

u - 1

乘以

8 u^{2} : 8 u^{3} - 8 u^{2}

将

8 u^{3} - 7 u - 1

减去

8 u^{3} - 8 u^{2}

得到新的余数

余数 = 8 u^{2} - 7 u - 1

因此

\frac{8 u ^{3} - 7 u - 1}{u - 1} = 8 u^{2} + \frac{8 u ^{2} - 7 u - 1}{u - 1}

= 8 u^{2} + \frac{8 u ^{2} - 7 u - 1}{u - 1}

对

\frac{8 u ^{2} - 7 u - 1}{u - 1}

做除法:

\frac{8 u ^{2} - 7 u - 1}{u - 1} = 8 u + \frac{u - 1}{u - 1}

将分子

8 u^{2} - 7 u - 1

与除数

u - 1

的首项系数相除：

\frac{8 u ^{2}}{u} = 8 u

商 = 8 u

将

u - 1

乘以

8 u : 8 u^{2} - 8 u

将

8 u^{2} - 7 u - 1

减去

8 u^{2} - 8 u

得到新的余数

余数 = u - 1

因此

\frac{8 u ^{2} - 7 u - 1}{u - 1} = 8 u + \frac{u - 1}{u - 1}

= 8 u^{2} + 8 u + \frac{u - 1}{u - 1}

对

\frac{u - 1}{u - 1}

做除法:

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

将分子

u - 1

与除数

u - 1

的首项系数相除：

\frac{u}{u} = 1

商 = 1

将

u - 1

乘以

1 : u - 1

将

u - 1

减去

u - 1

得到新的余数

余数 = 0

因此

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

= 8 u^{2} + 8 u + 1

= (u - 1) (8 u^{2} + 8 u + 1)

(u - 1) (8 u^{2} + 8 u + 1) = 0

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

u - 1 = 0 or 8 u^{2} + 8 u + 1 = 0

解

u - 1 = 0 : u = 1

u - 1 = 0

将

1

到右边

u - 1 = 0

两边加上

1

u - 1 + 1 = 0 + 1

化简

u = 1

u = 1

解

8 u^{2} + 8 u + 1 = 0 : u = \frac{- 2 + 2}{4}, u = - \frac{2 + 2}{4}

8 u^{2} + 8 u + 1 = 0

使用求根公式求解

8 u^{2} + 8 u + 1 = 0

二次方程求根公式:

若

a = 8, b = 8, c = 1

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 8 \cdot 1}{2 \cdot 8}

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 8 \cdot 1}{2 \cdot 8}

8^{2} - 4 \cdot 8 \cdot 1 = 42

8^{2} - 4 \cdot 8 \cdot 1

数字相乘:

4 \cdot 8 \cdot 1 = 32

= 8^{2} - 32

8^{2} = 64

= 64 - 32

数字相减:

64 - 32 = 32

= 32

32

质因数分解:

2^{5}

32

32

除以

232 = 16 \cdot 2

= 2 \cdot 16

16

除以

216 = 8 \cdot 2

= 2 \cdot 2 \cdot 8

8

除以

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

= 2^{5}

= 2^{5}

使用指数法则:

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

= 2^{4} \cdot 2

使用根式运算法则:

= 2 2^{4}

使用根式运算法则:

2^{4} = 2^{\frac{4}{2}} = 2^{2}

= 2^{2} 2

整理后得

= 42

u_{1, 2} = \frac{- 8 \pm 4 2}{2 \cdot 8}

将解分隔开

u_{1} = \frac{- 8 + 4 2}{2 \cdot 8}, u_{2} = \frac{- 8 - 4 2}{2 \cdot 8}

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

\frac{- 8 + 4 2}{2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{- 8 + 4 2}{16}

分解

- 8 + 42 : 4 (- 2 + 2)

- 8 + 42

改写为

= - 4 \cdot 2 + 42

因式分解出通项

4

= 4 (- 2 + 2)

= \frac{4 ( - 2 + 2 )}{16}

约分:

4

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

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

\frac{- 8 - 4 2}{2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{- 8 - 4 2}{16}

分解

- 8 - 42 : - 4 (2 + 2)

- 8 - 42

改写为

= - 4 \cdot 2 - 42

因式分解出通项

4

= - 4 (2 + 2)

= - \frac{4 ( 2 + 2 )}{16}

约分:

4

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

二次方程组的解是:

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

解为

u = 1, u = \frac{- 2 + 2}{4}, u = - \frac{2 + 2}{4}

u = cos (x)

代回

cos (x) = 1, cos (x) = \frac{- 2 + 2}{4}, cos (x) = - \frac{2 + 2}{4}

cos (x) = 1, cos (x) = \frac{- 2 + 2}{4}, cos (x) = - \frac{2 + 2}{4}

cos (x) = 1 : x = 2 πn

cos (x) = 1

cos (x) = 1

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 = 0 + 2 πn

x = 0 + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

cos (x) = \frac{- 2 + 2}{4} : x = arccos (\frac{- 2 + 2}{4}) + 2 πn, x = - arccos (\frac{- 2 + 2}{4}) + 2 πn

cos (x) = \frac{- 2 + 2}{4}

使用反三角函数性质

cos (x) = \frac{- 2 + 2}{4}

cos (x) = \frac{- 2 + 2}{4}

的通解

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

x = arccos (\frac{- 2 + 2}{4}) + 2 πn, x = - arccos (\frac{- 2 + 2}{4}) + 2 πn

x = arccos (\frac{- 2 + 2}{4}) + 2 πn, x = - arccos (\frac{- 2 + 2}{4}) + 2 πn

cos (x) = - \frac{2 + 2}{4} : x = arccos (- \frac{2 + 2}{4}) + 2 πn, x = - arccos (- \frac{2 + 2}{4}) + 2 πn

cos (x) = - \frac{2 + 2}{4}

使用反三角函数性质

cos (x) = - \frac{2 + 2}{4}

cos (x) = - \frac{2 + 2}{4}

的通解

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

x = arccos (- \frac{2 + 2}{4}) + 2 πn, x = - arccos (- \frac{2 + 2}{4}) + 2 πn

x = arccos (- \frac{2 + 2}{4}) + 2 πn, x = - arccos (- \frac{2 + 2}{4}) + 2 πn

合并所有解

x = 2 πn, x = arccos (\frac{- 2 + 2}{4}) + 2 πn, x = - arccos (\frac{- 2 + 2}{4}) + 2 πn, x = arccos (- \frac{2 + 2}{4}) + 2 πn, x = - arccos (- \frac{2 + 2}{4}) + 2 πn

以小数形式表示解

x = 2 πn, x = 1.71777 \dots + 2 πn, x = - 1.71777 \dots + 2 πn, x = 2.59356 \dots + 2 πn, x = - 2.59356 \dots + 2 πn

作图

流行的例子

sqrt(2)csc(x)-2=0

sin(2x)csc(x)=1,0<= x<= 2pi