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受欢迎的三角函数 >

20cos^6(x)-57cos^4(x)+27cos^2(x)=0

解答

20 cos^{6} (x) - 57 cos^{4} (x) + 27 cos^{2} (x) = 0

解答

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = 0.68471 \dots + 2 πn, x = 2 π - 0.68471 \dots + 2 πn, x = 2.45687 \dots + 2 πn, x = - 2.45687 \dots + 2 πn

+1

度数

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n, x = 39.23152 \dots^{\circ} + 36 0^{\circ} n, x = 320.76847 \dots^{\circ} + 36 0^{\circ} n, x = 140.76847 \dots^{\circ} + 36 0^{\circ} n, x = - 140.76847 \dots^{\circ} + 36 0^{\circ} n

求解步骤

20 cos^{6} (x) - 57 cos^{4} (x) + 27 cos^{2} (x) = 0

用替代法求解

20 cos^{6} (x) - 57 cos^{4} (x) + 27 cos^{2} (x) = 0

令

: cos (x) = u

20 u^{6} - 57 u^{4} + 27 u^{2} = 0

20 u^{6} - 57 u^{4} + 27 u^{2} = 0 : u = 0, u = \frac{3}{5}, u = - \frac{3}{5}, u = \frac{3}{2}, u = - \frac{3}{2}

20 u^{6} - 57 u^{4} + 27 u^{2} = 0

用

v = u^{2}, v^{2} = u^{4}

和

v^{3} = u^{6}

改写方程式

20 v^{3} - 57 v^{2} + 27 v = 0

解

20 v^{3} - 57 v^{2} + 27 v = 0 : v = 0, v = \frac{3}{5}, v = \frac{9}{4}

20 v^{3} - 57 v^{2} + 27 v = 0

因式分解

20 v^{3} - 57 v^{2} + 27 v : v (5 v - 3) (4 v - 9)

20 v^{3} - 57 v^{2} + 27 v

因式分解出通项

v : v (20 v^{2} - 57 v + 27)

20 v^{3} - 57 v^{2} + 27 v

使用指数法则:

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

v^{2} = vv

= 20 v^{2} v - 57 vv + 27 v

因式分解出通项

v

= v (20 v^{2} - 57 v + 27)

= v (20 v^{2} - 57 v + 27)

分解

20 v^{2} - 57 v + 27 : (5 v - 3) (4 v - 9)

20 v^{2} - 57 v + 27

将表达式拆分成组

20 v^{2} - 57 v + 27

定义

540

的因数:

1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 27, 30, 36, 45, 54, 60, 90, 108, 135, 180, 270, 540

540

约数 (因数)

找到

540

的质因数:

2, 2, 3, 3, 3, 5

540

540

除以

2540 = 270 \cdot 2

= 2 \cdot 270

270

除以

2270 = 135 \cdot 2

= 2 \cdot 2 \cdot 135

135

除以

3135 = 45 \cdot 3

= 2 \cdot 2 \cdot 3 \cdot 45

45

除以

345 = 15 \cdot 3

= 2 \cdot 2 \cdot 3 \cdot 3 \cdot 15

15

除以

315 = 5 \cdot 3

= 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 5

2, 3, 5

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 5

乘以

540

的质因数:

4, 6, 12, 9, 18, 36, 27, 54, 108, 10, 20, 15, 30, 60, 45, 90, 180, 135, 270

2 \cdot 2 = 4

2 \cdot 3 = 6

4, 6, 12, 9, 18, 36, 27, 54, 108, 10, 20, 15, 30, 60, 45, 90, 180, 135, 270

4, 6, 12, 9, 18, 36, 27, 54, 108, 10, 20, 15, 30, 60, 45, 90, 180, 135, 270

添加质因数:

2, 3, 5

将 1 和数字

540

自身相加

1, 540

540

的因数

1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 27, 30, 36, 45, 54, 60, 90, 108, 135, 180, 270, 540

540

的负因数:

- 1, - 2, - 3, - 4, - 5, - 6, - 9, - 10, - 12, - 15, - 18, - 20, - 27, - 30, - 36, - 45, - 54, - 60, - 90, - 108, - 135, - 180, - 270, - 540

将因数乘以

- 1

得到负因数

- 1, - 2, - 3, - 4, - 5, - 6, - 9, - 10, - 12, - 15, - 18, - 20, - 27, - 30, - 36, - 45, - 54, - 60, - 90, - 108, - 135, - 180, - 270, - 540

对于每两个因数

u * v = 540

，检验是否

u + v = - 57

检验

u = 1, v = 540 : u * v = 540, u + v = 541 \Rightarrow

假检验

u = 2, v = 270 : u * v = 540, u + v = 272 \Rightarrow

假

u = - 12, v = - 45

分组为

(a x^{2} + ux) + (vx + c)

(20 v^{2} - 12 v) + (- 45 v + 27)

= (20 v^{2} - 12 v) + (- 45 v + 27)

从

20 v^{2} - 12 v

分解出因式

4 v

:

4 v (5 v - 3)

20 v^{2} - 12 v

使用指数法则:

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

v^{2} = vv

= 20 vv - 12 v

将

12

改写为

4 \cdot 3

将

20

改写为

4 \cdot 5

= 4 \cdot 5 vv - 4 \cdot 3 v

因式分解出通项

4 v

= 4 v (5 v - 3)

从

- 45 v + 27

分解出因式

- 9

:

- 9 (5 v - 3)

- 45 v + 27

将

27

改写为

9 \cdot 3

将

45

改写为

9 \cdot 5

= - 9 \cdot 5 v + 9 \cdot 3

因式分解出通项

- 9

= - 9 (5 v - 3)

= 4 v (5 v - 3) - 9 (5 v - 3)

因式分解出通项

5 v - 3

= (5 v - 3) (4 v - 9)

= v (5 v - 3) (4 v - 9)

v (5 v - 3) (4 v - 9) = 0

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

v = 0 or 5 v - 3 = 0 or 4 v - 9 = 0

解

5 v - 3 = 0 : v = \frac{3}{5}

5 v - 3 = 0

将

3

到右边

5 v - 3 = 0

两边加上

3

5 v - 3 + 3 = 0 + 3

化简

5 v = 3

5 v = 3

两边除以

5

5 v = 3

两边除以

5

\frac{5 v}{5} = \frac{3}{5}

化简

v = \frac{3}{5}

v = \frac{3}{5}

解

4 v - 9 = 0 : v = \frac{9}{4}

4 v - 9 = 0

将

9

到右边

4 v - 9 = 0

两边加上

9

4 v - 9 + 9 = 0 + 9

化简

4 v = 9

4 v = 9

两边除以

4

4 v = 9

两边除以

4

\frac{4 v}{4} = \frac{9}{4}

化简

v = \frac{9}{4}

v = \frac{9}{4}

解为

v = 0, v = \frac{3}{5}, v = \frac{9}{4}

v = 0, v = \frac{3}{5}, v = \frac{9}{4}

代回

v = u^{2}

，求解

u

解

u^{2} = 0 : u = 0

u^{2} = 0

使用法则

x^{n} = 0 \Rightarrow x = 0

u = 0

解

u^{2} = \frac{3}{5} : u = \frac{3}{5}, u = - \frac{3}{5}

u^{2} = \frac{3}{5}

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = \frac{3}{5}, u = - \frac{3}{5}

解

u^{2} = \frac{9}{4} : u = \frac{3}{2}, u = - \frac{3}{2}

u^{2} = \frac{9}{4}

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = \frac{9}{4}, u = - \frac{9}{4}

\frac{9}{4} = \frac{3}{2}

\frac{9}{4}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{9}{4}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= \frac{9}{2}

9 = 3

9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 3

= \frac{3}{2}

- \frac{9}{4} = - \frac{3}{2}

- \frac{9}{4}

化简

\frac{9}{4} : \frac{3}{2}

\frac{9}{4}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{9}{4}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= \frac{9}{2}

9 = 3

9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 3

= \frac{3}{2}

= - \frac{3}{2}

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

解为

u = 0, u = \frac{3}{5}, u = - \frac{3}{5}, u = \frac{3}{2}, u = - \frac{3}{2}

u = cos (x)

代回

cos (x) = 0, cos (x) = \frac{3}{5}, cos (x) = - \frac{3}{5}, cos (x) = \frac{3}{2}, cos (x) = - \frac{3}{2}

cos (x) = 0, cos (x) = \frac{3}{5}, cos (x) = - \frac{3}{5}, cos (x) = \frac{3}{2}, cos (x) = - \frac{3}{2}

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

cos (x) = 0

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}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

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

cos (x) = \frac{3}{5} : x = arccos (\frac{3}{5}) + 2 πn, x = 2 π - arccos (\frac{3}{5}) + 2 πn

cos (x) = \frac{3}{5}

使用反三角函数性质

cos (x) = \frac{3}{5}

cos (x) = \frac{3}{5}

的通解

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

x = arccos (\frac{3}{5}) + 2 πn, x = 2 π - arccos (\frac{3}{5}) + 2 πn

x = arccos (\frac{3}{5}) + 2 πn, x = 2 π - arccos (\frac{3}{5}) + 2 πn

cos (x) = - \frac{3}{5} : x = arccos (- \frac{3}{5}) + 2 πn, x = - arccos (- \frac{3}{5}) + 2 πn

cos (x) = - \frac{3}{5}

使用反三角函数性质

cos (x) = - \frac{3}{5}

cos (x) = - \frac{3}{5}

的通解

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

x = arccos (- \frac{3}{5}) + 2 πn, x = - arccos (- \frac{3}{5}) + 2 πn

x = arccos (- \frac{3}{5}) + 2 πn, x = - arccos (- \frac{3}{5}) + 2 πn

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

无解

cos (x) = \frac{3}{2}

- 1 \leq cos (x) \leq 1

无解

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

无解

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

- 1 \leq cos (x) \leq 1

无解

合并所有解

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = arccos (\frac{3}{5}) + 2 πn, x = 2 π - arccos (\frac{3}{5}) + 2 πn, x = arccos (- \frac{3}{5}) + 2 πn, x = - arccos (- \frac{3}{5}) + 2 πn

以小数形式表示解

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = 0.68471 \dots + 2 πn, x = 2 π - 0.68471 \dots + 2 πn, x = 2.45687 \dots + 2 πn, x = - 2.45687 \dots + 2 πn

作图

流行的例子

2sin^2(w)-3sin(w)+1=0

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

2+sin(x)=5sin(x)

2 + sin (x) = 5 sin (x)

2 cos (θ) + 5 = 0

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

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

2cos(2θ)+1=0,0<= θ<= 2pi

2 cos (2 θ) + 1 = 0, 0 \leq θ \leq 2 π