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 >

2cos(x)+2cos(2x)=0

解答

2 cos (x) + 2 cos (2 x) = 0

解答

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

+1

度数

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

2 cos (x) + 2 cos (2 x) = 0

使用三角恒等式改写

2 cos (2 x) + 2 cos (x)

使用倍角公式:

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

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

(- 1 + 2 cos^{2} (x)) \cdot 2 + 2 cos (x) = 0

用替代法求解

(- 1 + 2 cos^{2} (x)) \cdot 2 + 2 cos (x) = 0

令

: cos (x) = u

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

(- 1 + 2 u^{2}) \cdot 2 + 2 u = 0 : u = \frac{1}{2}, u = - 1

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

展开

(- 1 + 2 u^{2}) \cdot 2 + 2 u : - 2 + 4 u^{2} + 2 u

(- 1 + 2 u^{2}) \cdot 2 + 2 u

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

乘开

2 (- 1 + 2 u^{2}) : - 2 + 4 u^{2}

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

使用分配律:

a (b + c) = ab + a c

a = 2, b = - 1, c = 2 u^{2}

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

使用加减运算法则

+ (- a) = - a

= - 2 \cdot 1 + 2 \cdot 2 u^{2}

化简

- 2 \cdot 1 + 2 \cdot 2 u^{2} : - 2 + 4 u^{2}

- 2 \cdot 1 + 2 \cdot 2 u^{2}

数字相乘:

2 \cdot 1 = 2

= - 2 + 2 \cdot 2 u^{2}

数字相乘:

2 \cdot 2 = 4

= - 2 + 4 u^{2}

= - 2 + 4 u^{2}

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = 4, b = 2, c = - 2

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

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

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

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

使用法则

- (- a) = a

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

数字相乘:

4 \cdot 4 \cdot 2 = 32

= 2^{2} + 32

2^{2} = 4

= 4 + 32

数字相加:

4 + 32 = 36

= 36

因式分解数字:

36 = 6^{2}

= 6^{2}

使用根式运算法则:

n a^{n} = a

6^{2} = 6

= 6

u_{1, 2} = \frac{- 2 \pm 6}{2 \cdot 4}

将解分隔开

u_{1} = \frac{- 2 + 6}{2 \cdot 4}, u_{2} = \frac{- 2 - 6}{2 \cdot 4}

u = \frac{- 2 + 6}{2 \cdot 4} : \frac{1}{2}

\frac{- 2 + 6}{2 \cdot 4}

数字相加/相减:

- 2 + 6 = 4

= \frac{4}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{4}{8}

约分:

4

= \frac{1}{2}

u = \frac{- 2 - 6}{2 \cdot 4} : - 1

\frac{- 2 - 6}{2 \cdot 4}

数字相减:

- 2 - 6 = - 8

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

数字相乘:

2 \cdot 4 = 8

= \frac{- 8}{8}

使用分式法则:

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

= - \frac{8}{8}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

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

u = cos (x)

代回

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

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

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) = - 1 : x = π + 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 = π + 2 πn

x = π + 2 πn

合并所有解

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

作图

流行的例子

cos^2(x)+sin(x)=0

cos^{2} (x) + sin (x) = 0

tan(x)=(-sqrt(3))/3

tan (x) = \frac{- 3}{3}

tan(x/3)=(sqrt(3))/3

tan (\frac{x}{3}) = \frac{3}{3}

12sin^2(x)+sin(x)-1=0

12 sin^{2} (x) + sin (x) - 1 = 0

2 sin^{2} (2 x) - 1 = 0