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 >

1=sqrt(4cos(2θ))

解答

1 = 4 cos (2 θ)

解答

θ = \frac{1.31811 \dots}{2} + πn, θ = π - \frac{1.31811 \dots}{2} + πn

+1

度数

θ = 37.76124 \dots^{\circ} + 18 0^{\circ} n, θ = 142.23875 \dots^{\circ} + 18 0^{\circ} n

求解步骤

1 = 4 cos (2 θ)

交换两边

4 cos (2 θ) = 1

两边进行平方:

4 cos (2 θ) = 1

4 cos (2 θ) = 1

(4 cos (2 θ))^{2} = 1^{2}

展开

(4 cos (2 θ))^{2} : 4 cos (2 θ)

(4 cos (2 θ))^{2}

使用根式运算法则:

a = a^{\frac{1}{2}}

= ((4 cos (2 θ))^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= (4 cos (2 θ))^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 4 cos (2 θ)

展开

1^{2} : 1

1^{2}

使用法则

1^{a} = 1

= 1

4 cos (2 θ) = 1

4 cos (2 θ) = 1

解

4 cos (2 θ) = 1 : cos (2 θ) = \frac{1}{4}

4 cos (2 θ) = 1

两边除以

4

4 cos (2 θ) = 1

两边除以

4

\frac{4 cos ( 2 θ )}{4} = \frac{1}{4}

化简

cos (2 θ) = \frac{1}{4}

cos (2 θ) = \frac{1}{4}

cos (2 θ) = \frac{1}{4}

验证解:

cos (2 θ) = \frac{1}{4}

真

将它们代入

4 cos (2 θ) = 1

检验解是否符合

去除与方程不符的解。

代入

cos (2 θ) = \frac{1}{4} :

真

4 (\frac{1}{4}) = 1

4 (\frac{1}{4}) = 1

4 (\frac{1}{4})

去除括号:

(a) = a

= 4 \cdot \frac{1}{4}

乘

4 \cdot \frac{1}{4} : 1

4 \cdot \frac{1}{4}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 4}{4}

约分:

4

= 1

= 1

使用法则

1 = 1

= 1

1 = 1

真

解是

cos (2 θ) = \frac{1}{4}

使用反三角函数性质

cos (2 θ) = \frac{1}{4}

cos (2 θ) = \frac{1}{4}

的通解

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

2 θ = arccos (\frac{1}{4}) + 2 πn, 2 θ = 2 π - arccos (\frac{1}{4}) + 2 πn

2 θ = arccos (\frac{1}{4}) + 2 πn, 2 θ = 2 π - arccos (\frac{1}{4}) + 2 πn

解

2 θ = arccos (\frac{1}{4}) + 2 πn : θ = \frac{arccos ( \frac{1}{4} )}{2} + πn

2 θ = arccos (\frac{1}{4}) + 2 πn

两边除以

2

2 θ = arccos (\frac{1}{4}) + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{arccos ( \frac{1}{4} )}{2} + \frac{2 πn}{2}

化简

θ = \frac{arccos ( \frac{1}{4} )}{2} + πn

θ = \frac{arccos ( \frac{1}{4} )}{2} + πn

解

2 θ = 2 π - arccos (\frac{1}{4}) + 2 πn : θ = π - \frac{arccos ( \frac{1}{4} )}{2} + πn

2 θ = 2 π - arccos (\frac{1}{4}) + 2 πn

两边除以

2

2 θ = 2 π - arccos (\frac{1}{4}) + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{2 π}{2} - \frac{arccos ( \frac{1}{4} )}{2} + \frac{2 πn}{2}

化简

θ = π - \frac{arccos ( \frac{1}{4} )}{2} + πn

θ = π - \frac{arccos ( \frac{1}{4} )}{2} + πn

θ = \frac{arccos ( \frac{1}{4} )}{2} + πn, θ = π - \frac{arccos ( \frac{1}{4} )}{2} + πn

以小数形式表示解

θ = \frac{1.31811 \dots}{2} + πn, θ = π - \frac{1.31811 \dots}{2} + πn

作图

流行的例子

209cos(47)=150cos(θ)

209 cos (4 7^{\circ}) = 150 cos (θ)

7-5cos(x)=10,7-5cos(x)=10

7 - 5 cos (x) = 10, 7 - 5 cos (x) = 10

cos (4 x) = cos (x)

-4cos(2θ)-19=-26cos(θ),0<= θ<360

- 4 cos (2 θ) - 19 = - 26 cos (θ), 0 \leq θ < 36 0^{\circ}

3cos^2(x)=7-8sin(x)

3 cos^{2} (x) = 7 - 8 sin (x)