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受欢迎的三角函数 >

6cos^3(x)+cos^2(x)-1=0

解答

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

解答

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

+1

度数

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

求解步骤

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

用替代法求解

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

令

: cos (x) = u

6 u^{3} + u^{2} - 1 = 0

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

6 u^{3} + u^{2} - 1 = 0

因式分解

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

6 u^{3} + u^{2} - 1

使用有理根定理

a_{0} = 1, a_{n} = 6

a_{0}

的除数

: 1, a_{n}

的除数

: 1, 2, 3, 6

因此，检验以下有理数:

\pm \frac{1}{1 , 2 , 3 , 6}

\frac{1}{2}

是表达式的根，所以因式分解

2 u - 1

= (2 u - 1) \frac{6 u ^{3} + u ^{2} - 1}{2 u - 1}

\frac{6 u ^{3} + u ^{2} - 1}{2 u - 1} = 3 u^{2} + 2 u + 1

\frac{6 u ^{3} + u ^{2} - 1}{2 u - 1}

对

\frac{6 u ^{3} + u ^{2} - 1}{2 u - 1}

做除法:

\frac{6 u ^{3} + u ^{2} - 1}{2 u - 1} = 3 u^{2} + \frac{4 u ^{2} - 1}{2 u - 1}

将分子

6 u^{3} + u^{2} - 1

与除数

2 u - 1

的首项系数相除：

\frac{6 u ^{3}}{2 u} = 3 u^{2}

商 = 3 u^{2}

将

2 u - 1

乘以

3 u^{2} : 6 u^{3} - 3 u^{2}

将

6 u^{3} + u^{2} - 1

减去

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

得到新的余数

余数 = 4 u^{2} - 1

因此

\frac{6 u ^{3} + u ^{2} - 1}{2 u - 1} = 3 u^{2} + \frac{4 u ^{2} - 1}{2 u - 1}

= 3 u^{2} + \frac{4 u ^{2} - 1}{2 u - 1}

对

\frac{4 u ^{2} - 1}{2 u - 1}

做除法:

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

将分子

4 u^{2} - 1

与除数

2 u - 1

的首项系数相除：

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

商 = 2 u

将

2 u - 1

乘以

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

将

4 u^{2} - 1

减去

4 u^{2} - 2 u

得到新的余数

余数 = 2 u - 1

因此

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

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

对

\frac{2 u - 1}{2 u - 1}

做除法:

\frac{2 u - 1}{2 u - 1} = 1

将分子

2 u - 1

与除数

2 u - 1

的首项系数相除：

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

商 = 1

将

2 u - 1

乘以

1 : 2 u - 1

将

2 u - 1

减去

2 u - 1

得到新的余数

余数 = 0

因此

\frac{2 u - 1}{2 u - 1} = 1

= 3 u^{2} + 2 u + 1

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

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

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

2 u - 1 = 0 or 3 u^{2} + 2 u + 1 = 0

解

2 u - 1 = 0 : u = \frac{1}{2}

2 u - 1 = 0

将

1

到右边

2 u - 1 = 0

两边加上

1

2 u - 1 + 1 = 0 + 1

化简

2 u = 1

2 u = 1

两边除以

2

2 u = 1

两边除以

2

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

化简

u = \frac{1}{2}

u = \frac{1}{2}

解

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = 3, b = 2, c = 1

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

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

化简

2^{2} - 4 \cdot 3 \cdot 1 : 22 i

2^{2} - 4 \cdot 3 \cdot 1

数字相乘:

4 \cdot 3 \cdot 1 = 12

= 2^{2} - 12

使用虚数运算法则:

- a = i a

= i 12 - 2^{2}

- 2^{2} + 12 = 22

- 2^{2} + 12

2^{2} = 4

= - 4 + 12

数字相加/相减:

- 4 + 12 = 8

= 8

8

质因数分解:

2^{3}

8

8

除以

28 = 4 \cdot 2

= 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 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

= 2^{3}

= 2^{3}

使用指数法则:

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

= 2^{2} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 22

= 22 i

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

将解分隔开

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

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

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

数字相乘:

2 \cdot 3 = 6

= \frac{- 2 + 2 2 i}{6}

分解

- 2 + 22 i : 2 (- 1 + 2 i)

- 2 + 22 i

改写为

= - 2 \cdot 1 + 22 i

因式分解出通项

2

= 2 (- 1 + 2 i)

= \frac{2 ( - 1 + 2 i )}{6}

约分:

2

= \frac{- 1 + 2 i}{3}

将

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

改写成标准复数形式:

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

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

使用分式法则:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{- 1 + 2 i}{3} = - \frac{1}{3} + \frac{2 i}{3}

= - \frac{1}{3} + \frac{2 i}{3}

= - \frac{1}{3} + \frac{2}{3} i

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

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

数字相乘:

2 \cdot 3 = 6

= \frac{- 2 - 2 2 i}{6}

分解

- 2 - 22 i : - 2 (1 + 2 i)

- 2 - 22 i

改写为

= - 2 \cdot 1 - 22 i

因式分解出通项

2

= - 2 (1 + 2 i)

= - \frac{2 ( 1 + 2 i )}{6}

约分:

2

= - \frac{1 + 2 i}{3}

将

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

改写成标准复数形式:

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

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

使用分式法则:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

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

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

去除括号:

(a) = a

= - \frac{1}{3} - \frac{2 i}{3}

= - \frac{1}{3} - \frac{2}{3} i

二次方程组的解是:

u = - \frac{1}{3} + i \frac{2}{3}, u = - \frac{1}{3} - i \frac{2}{3}

解为

u = \frac{1}{2}, u = - \frac{1}{3} + i \frac{2}{3}, u = - \frac{1}{3} - i \frac{2}{3}

u = cos (x)

代回

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

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

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

cos (x) = - \frac{1}{3} + i \frac{2}{3} :

无解

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

无解

cos (x) = - \frac{1}{3} - i \frac{2}{3} :

无解

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

无解

合并所有解

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

作图

流行的例子

4tan^2(x)+12tan(x)-27=0

4 tan^{2} (x) + 12 tan (x) - 27 = 0

sin^2(x)-cos(x)= 1/4

sin^{2} (x) - cos (x) = \frac{1}{4}

cos^4(a)=3+4cos^2(a)+cos^4(a)

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

cos^{4} (t) = 1

8cos^2(x)-12sin(x)-12=0

8 cos^{2} (x) - 12 sin (x) - 12 = 0