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 >

5cos(2x)-cos(x)+3=0

解答

5 cos (2 x) - cos (x) + 3 = 0

解答

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = 1.98231 \dots + 2 πn, x = - 1.98231 \dots + 2 πn

+1

度数

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n, x = 113.57817 \dots^{\circ} + 36 0^{\circ} n, x = - 113.57817 \dots^{\circ} + 36 0^{\circ} n

求解步骤

5 cos (2 x) - cos (x) + 3 = 0

使用三角恒等式改写

3 - cos (x) + 5 cos (2 x)

使用倍角公式:

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

= 3 - cos (x) + 5 (2 cos^{2} (x) - 1)

化简

3 - cos (x) + 5 (2 cos^{2} (x) - 1) : 10 cos^{2} (x) - cos (x) - 2

3 - cos (x) + 5 (2 cos^{2} (x) - 1)

乘开

5 (2 cos^{2} (x) - 1) : 10 cos^{2} (x) - 5

5 (2 cos^{2} (x) - 1)

使用分配律:

a (b - c) = ab - a c

a = 5, b = 2 cos^{2} (x), c = 1

= 5 \cdot 2 cos^{2} (x) - 5 \cdot 1

化简

5 \cdot 2 cos^{2} (x) - 5 \cdot 1 : 10 cos^{2} (x) - 5

5 \cdot 2 cos^{2} (x) - 5 \cdot 1

数字相乘:

5 \cdot 2 = 10

= 10 cos^{2} (x) - 5 \cdot 1

数字相乘:

5 \cdot 1 = 5

= 10 cos^{2} (x) - 5

= 10 cos^{2} (x) - 5

= 3 - cos (x) + 10 cos^{2} (x) - 5

化简

3 - cos (x) + 10 cos^{2} (x) - 5 : 10 cos^{2} (x) - cos (x) - 2

3 - cos (x) + 10 cos^{2} (x) - 5

对同类项分组

= - cos (x) + 10 cos^{2} (x) + 3 - 5

数字相加/相减:

3 - 5 = - 2

= 10 cos^{2} (x) - cos (x) - 2

= 10 cos^{2} (x) - cos (x) - 2

= 10 cos^{2} (x) - cos (x) - 2

- 2 - cos (x) + 10 cos^{2} (x) = 0

用替代法求解

- 2 - cos (x) + 10 cos^{2} (x) = 0

令

: cos (x) = u

- 2 - u + 10 u^{2} = 0

- 2 - u + 10 u^{2} = 0 : u = \frac{1}{2}, u = - \frac{2}{5}

- 2 - u + 10 u^{2} = 0

改写成标准形式

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

10 u^{2} - u - 2 = 0

使用求根公式求解

10 u^{2} - u - 2 = 0

二次方程求根公式:

若

a = 10, b = - 1, c = - 2

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 10 ( - 2 )}{2 \cdot 10}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 10 ( - 2 )}{2 \cdot 10}

(- 1)^{2} - 4 \cdot 10 (- 2) = 9

(- 1)^{2} - 4 \cdot 10 (- 2)

使用法则

- (- a) = a

= (- 1)^{2} + 4 \cdot 10 \cdot 2

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

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

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 10 \cdot 2 = 80

4 \cdot 10 \cdot 2

数字相乘:

4 \cdot 10 \cdot 2 = 80

= 80

= 1 + 80

数字相加:

1 + 80 = 81

= 81

因式分解数字:

81 = 9^{2}

= 9^{2}

使用根式运算法则:

n a^{n} = a

9^{2} = 9

= 9

u_{1, 2} = \frac{- ( - 1 ) \pm 9}{2 \cdot 10}

将解分隔开

u_{1} = \frac{- ( - 1 ) + 9}{2 \cdot 10}, u_{2} = \frac{- ( - 1 ) - 9}{2 \cdot 10}

u = \frac{- ( - 1 ) + 9}{2 \cdot 10} : \frac{1}{2}

\frac{- ( - 1 ) + 9}{2 \cdot 10}

使用法则

- (- a) = a

= \frac{1 + 9}{2 \cdot 10}

数字相加:

1 + 9 = 10

= \frac{10}{2 \cdot 10}

数字相乘:

2 \cdot 10 = 20

= \frac{10}{20}

约分:

10

= \frac{1}{2}

u = \frac{- ( - 1 ) - 9}{2 \cdot 10} : - \frac{2}{5}

\frac{- ( - 1 ) - 9}{2 \cdot 10}

使用法则

- (- a) = a

= \frac{1 - 9}{2 \cdot 10}

数字相减:

1 - 9 = - 8

= \frac{- 8}{2 \cdot 10}

数字相乘:

2 \cdot 10 = 20

= \frac{- 8}{20}

使用分式法则:

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

= - \frac{8}{20}

约分:

4

= - \frac{2}{5}

二次方程组的解是:

u = \frac{1}{2}, u = - \frac{2}{5}

u = cos (x)

代回

cos (x) = \frac{1}{2}, cos (x) = - \frac{2}{5}

cos (x) = \frac{1}{2}, cos (x) = - \frac{2}{5}

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

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

使用反三角函数性质

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

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

的通解

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

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

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

合并所有解

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

以小数形式表示解

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = 1.98231 \dots + 2 πn, x = - 1.98231 \dots + 2 πn

作图

流行的例子

4tan^2(θ)=12,0<= θ<2pi

4 tan^{2} (θ) = 12, 0 \leq θ < 2 π

(1+tanh(x))/(1-tanh(x))=2

\frac{1 + tanh ( x )}{1 - tanh ( x )} = 2

sin(2x)-sin(x)=0,-pi<= x<= pi

sin (2 x) - sin (x) = 0, - π \leq x \leq π

2(tan(x)+3)=5+tan(x)

2 (tan (x) + 3) = 5 + tan (x)

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