zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

4cos(2x)=2-8cos(x)

解答

4 cos (2 x) = 2 - 8 cos (x)

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

4 cos (2 x) = 2 - 8 cos (x)

两边减去

2 - 8 cos (x)

4 cos (2 x) - 2 + 8 cos (x) = 0

使用三角恒等式改写

- 2 + 4 cos (2 x) + 8 cos (x)

使用倍角公式:

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

= - 2 + 4 (2 cos^{2} (x) - 1) + 8 cos (x)

化简

- 2 + 4 (2 cos^{2} (x) - 1) + 8 cos (x) : 8 cos^{2} (x) + 8 cos (x) - 6

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

乘开

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

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

使用分配律:

a (b - c) = ab - a c

a = 4, b = 2 cos^{2} (x), c = 1

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

化简

4 \cdot 2 cos^{2} (x) - 4 \cdot 1 : 8 cos^{2} (x) - 4

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

数字相乘:

4 \cdot 2 = 8

= 8 cos^{2} (x) - 4 \cdot 1

数字相乘:

4 \cdot 1 = 4

= 8 cos^{2} (x) - 4

= 8 cos^{2} (x) - 4

= - 2 + 8 cos^{2} (x) - 4 + 8 cos (x)

化简

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

- 2 + 8 cos^{2} (x) - 4 + 8 cos (x)

对同类项分组

= 8 cos^{2} (x) + 8 cos (x) - 2 - 4

数字相减:

- 2 - 4 = - 6

= 8 cos^{2} (x) + 8 cos (x) - 6

= 8 cos^{2} (x) + 8 cos (x) - 6

= 8 cos^{2} (x) + 8 cos (x) - 6

- 6 + 8 cos (x) + 8 cos^{2} (x) = 0

用替代法求解

- 6 + 8 cos (x) + 8 cos^{2} (x) = 0

令

: cos (x) = u

- 6 + 8 u + 8 u^{2} = 0

- 6 + 8 u + 8 u^{2} = 0 : u = \frac{1}{2}, u = - \frac{3}{2}

- 6 + 8 u + 8 u^{2} = 0

改写成标准形式

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

8 u^{2} + 8 u - 6 = 0

使用求根公式求解

8 u^{2} + 8 u - 6 = 0

二次方程求根公式:

若

a = 8, b = 8, c = - 6

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 8 ( - 6 )}{2 \cdot 8}

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 8 ( - 6 )}{2 \cdot 8}

8^{2} - 4 \cdot 8 (- 6) = 16

8^{2} - 4 \cdot 8 (- 6)

使用法则

- (- a) = a

= 8^{2} + 4 \cdot 8 \cdot 6

数字相乘:

4 \cdot 8 \cdot 6 = 192

= 8^{2} + 192

8^{2} = 64

= 64 + 192

数字相加:

64 + 192 = 256

= 256

因式分解数字:

256 = 1 6^{2}

= 1 6^{2}

使用根式运算法则:

n a^{n} = a

1 6^{2} = 16

= 16

u_{1, 2} = \frac{- 8 \pm 16}{2 \cdot 8}

将解分隔开

u_{1} = \frac{- 8 + 16}{2 \cdot 8}, u_{2} = \frac{- 8 - 16}{2 \cdot 8}

u = \frac{- 8 + 16}{2 \cdot 8} : \frac{1}{2}

\frac{- 8 + 16}{2 \cdot 8}

数字相加/相减:

- 8 + 16 = 8

= \frac{8}{2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{8}{16}

约分:

8

= \frac{1}{2}

u = \frac{- 8 - 16}{2 \cdot 8} : - \frac{3}{2}

\frac{- 8 - 16}{2 \cdot 8}

数字相减:

- 8 - 16 = - 24

= \frac{- 24}{2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{- 24}{16}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{24}{16}

约分:

8

= - \frac{3}{2}

二次方程组的解是:

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

u = cos (x)

代回

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

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

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

作图

流行的例子

tan (x) = \frac{7}{8}

4 cos^{2} (x) + 1 = 0

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

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

cos (x - \frac{π}{2}) = 0

2csc^2(2x)-csc(2x)-6=0

2 csc^{2} (2 x) - csc (2 x) - 6 = 0