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(2θ)=-5cos(θ)

解答

5 cos (2 θ) = - 5 cos (θ)

解答

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

+1

度数

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

求解步骤

5 cos (2 θ) = - 5 cos (θ)

两边减去

- 5 cos (θ)

5 cos (2 θ) + 5 cos (θ) = 0

使用三角恒等式改写

5 cos (2 θ) + 5 cos (θ)

使用倍角公式:

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

= 5 (2 cos^{2} (θ) - 1) + 5 cos (θ)

(- 1 + 2 cos^{2} (θ)) \cdot 5 + 5 cos (θ) = 0

用替代法求解

(- 1 + 2 cos^{2} (θ)) \cdot 5 + 5 cos (θ) = 0

令

: cos (θ) = u

(- 1 + 2 u^{2}) \cdot 5 + 5 u = 0

(- 1 + 2 u^{2}) \cdot 5 + 5 u = 0 : u = \frac{1}{2}, u = - 1

(- 1 + 2 u^{2}) \cdot 5 + 5 u = 0

展开

(- 1 + 2 u^{2}) \cdot 5 + 5 u : - 5 + 10 u^{2} + 5 u

(- 1 + 2 u^{2}) \cdot 5 + 5 u

= 5 (- 1 + 2 u^{2}) + 5 u

乘开

5 (- 1 + 2 u^{2}) : - 5 + 10 u^{2}

5 (- 1 + 2 u^{2})

使用分配律:

a (b + c) = ab + a c

a = 5, b = - 1, c = 2 u^{2}

= 5 (- 1) + 5 \cdot 2 u^{2}

使用加减运算法则

+ (- a) = - a

= - 5 \cdot 1 + 5 \cdot 2 u^{2}

化简

- 5 \cdot 1 + 5 \cdot 2 u^{2} : - 5 + 10 u^{2}

- 5 \cdot 1 + 5 \cdot 2 u^{2}

数字相乘:

5 \cdot 1 = 5

= - 5 + 5 \cdot 2 u^{2}

数字相乘:

5 \cdot 2 = 10

= - 5 + 10 u^{2}

= - 5 + 10 u^{2}

= - 5 + 10 u^{2} + 5 u

- 5 + 10 u^{2} + 5 u = 0

改写成标准形式

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

10 u^{2} + 5 u - 5 = 0

使用求根公式求解

10 u^{2} + 5 u - 5 = 0

二次方程求根公式:

若

a = 10, b = 5, c = - 5

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 10 ( - 5 )}{2 \cdot 10}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 10 ( - 5 )}{2 \cdot 10}

5^{2} - 4 \cdot 10 (- 5) = 15

5^{2} - 4 \cdot 10 (- 5)

使用法则

- (- a) = a

= 5^{2} + 4 \cdot 10 \cdot 5

数字相乘:

4 \cdot 10 \cdot 5 = 200

= 5^{2} + 200

5^{2} = 25

= 25 + 200

数字相加:

25 + 200 = 225

= 225

因式分解数字:

225 = 1 5^{2}

= 1 5^{2}

使用根式运算法则:

n a^{n} = a

1 5^{2} = 15

= 15

u_{1, 2} = \frac{- 5 \pm 15}{2 \cdot 10}

将解分隔开

u_{1} = \frac{- 5 + 15}{2 \cdot 10}, u_{2} = \frac{- 5 - 15}{2 \cdot 10}

u = \frac{- 5 + 15}{2 \cdot 10} : \frac{1}{2}

\frac{- 5 + 15}{2 \cdot 10}

数字相加/相减:

- 5 + 15 = 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{- 5 - 15}{2 \cdot 10} : - 1

\frac{- 5 - 15}{2 \cdot 10}

数字相减:

- 5 - 15 = - 20

= \frac{- 20}{2 \cdot 10}

数字相乘:

2 \cdot 10 = 20

= \frac{- 20}{20}

使用分式法则:

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

= - \frac{20}{20}

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 (θ)

代回

cos (θ) = \frac{1}{2}, cos (θ) = - 1

cos (θ) = \frac{1}{2}, cos (θ) = - 1

cos (θ) = \frac{1}{2} : θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

cos (θ) = \frac{1}{2}

cos (θ) = \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}

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

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

cos (θ) = - 1 : θ = π + 2 πn

cos (θ) = - 1

cos (θ) = - 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}

θ = π + 2 πn

θ = π + 2 πn

合并所有解

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

作图

流行的例子

sin(θ)= 3/5 ,0<θ< pi/2

sin (θ) = \frac{3}{5}, 0 < θ < \frac{π}{2}

cos(4x)cos(x)+sin(4x)sin(x)= 1/2

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

tan^2(θ)+7tan(θ)+5=0

tan^{2} (θ) + 7 tan (θ) + 5 = 0

tan(θ)=(2sqrt(3))/1

tan (θ) = \frac{2 3}{1}

cos^2(x)-cos(x)=0,x,0<= x<= 2pi

cos^{2} (x) - cos (x) = 0, x, 0 \leq x \leq 2 π