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^2(4x)-1=0

解答

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

解答

x = \frac{π}{16} + \frac{πn}{2}, x = \frac{π}{2} - \frac{π}{16} + \frac{πn}{2}, x = \frac{3 π}{16} + \frac{πn}{2}, x = - \frac{3 π}{16} + \frac{πn}{2}

+1

度数

x = 11.2 5^{\circ} + 9 0^{\circ} n, x = 78.7 5^{\circ} + 9 0^{\circ} n, x = 33.7 5^{\circ} + 9 0^{\circ} n, x = - 33.7 5^{\circ} + 9 0^{\circ} n

求解步骤

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

用替代法求解

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

令

: cos (4 x) = u

2 u^{2} - 1 = 0

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

2 u^{2} - 1 = 0

将

1

到右边

2 u^{2} - 1 = 0

两边加上

1

2 u^{2} - 1 + 1 = 0 + 1

化简

2 u^{2} = 1

2 u^{2} = 1

两边除以

2

2 u^{2} = 1

两边除以

2

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

化简

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

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

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

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

u = cos (4 x)

代回

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

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

cos (4 x) = \frac{1}{2} : x = \frac{π}{16} + \frac{πn}{2}, x = \frac{π}{2} - \frac{π}{16} + \frac{πn}{2}

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

使用反三角函数性质

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

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

的通解

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

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

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

解

4 x = arccos (\frac{1}{2}) + 2 πn : x = \frac{π}{16} + \frac{πn}{2}

4 x = arccos (\frac{1}{2}) + 2 πn

化简

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

arccos (\frac{1}{2}) + 2 πn

使用以下普通恒等式:

arccos (\frac{1}{2}) = \frac{π}{4}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{4} + 2 πn

4 x = \frac{π}{4} + 2 πn

两边除以

4

4 x = \frac{π}{4} + 2 πn

两边除以

4

\frac{4 x}{4} = \frac{\frac{π}{4}}{4} + \frac{2 πn}{4}

化简

\frac{4 x}{4} = \frac{\frac{π}{4}}{4} + \frac{2 πn}{4}

化简

\frac{4 x}{4} : x

\frac{4 x}{4}

数字相除:

\frac{4}{4} = 1

= x

化简

\frac{\frac{π}{4}}{4} + \frac{2 πn}{4} : \frac{π}{16} + \frac{πn}{2}

\frac{\frac{π}{4}}{4} + \frac{2 πn}{4}

\frac{\frac{π}{4}}{4} = \frac{π}{16}

\frac{\frac{π}{4}}{4}

使用分式法则:

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

= \frac{π}{4 \cdot 4}

数字相乘:

4 \cdot 4 = 16

= \frac{π}{16}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

约分:

2

= \frac{πn}{2}

= \frac{π}{16} + \frac{πn}{2}

x = \frac{π}{16} + \frac{πn}{2}

x = \frac{π}{16} + \frac{πn}{2}

x = \frac{π}{16} + \frac{πn}{2}

解

4 x = 2 π - arccos (\frac{1}{2}) + 2 πn : x = \frac{π}{2} - \frac{π}{16} + \frac{πn}{2}

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

化简

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

2 π - arccos (\frac{1}{2}) + 2 πn

使用以下普通恒等式:

arccos (\frac{1}{2}) = \frac{π}{4}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= 2 π - \frac{π}{4} + 2 πn

4 x = 2 π - \frac{π}{4} + 2 πn

两边除以

4

4 x = 2 π - \frac{π}{4} + 2 πn

两边除以

4

\frac{4 x}{4} = \frac{2 π}{4} - \frac{\frac{π}{4}}{4} + \frac{2 πn}{4}

化简

\frac{4 x}{4} = \frac{2 π}{4} - \frac{\frac{π}{4}}{4} + \frac{2 πn}{4}

化简

\frac{4 x}{4} : x

\frac{4 x}{4}

数字相除:

\frac{4}{4} = 1

= x

化简

\frac{2 π}{4} - \frac{\frac{π}{4}}{4} + \frac{2 πn}{4} : \frac{π}{2} - \frac{π}{16} + \frac{πn}{2}

\frac{2 π}{4} - \frac{\frac{π}{4}}{4} + \frac{2 πn}{4}

\frac{2 π}{4} = \frac{π}{2}

\frac{2 π}{4}

约分:

2

= \frac{π}{2}

\frac{\frac{π}{4}}{4} = \frac{π}{16}

\frac{\frac{π}{4}}{4}

使用分式法则:

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

= \frac{π}{4 \cdot 4}

数字相乘:

4 \cdot 4 = 16

= \frac{π}{16}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

约分:

2

= \frac{πn}{2}

= \frac{π}{2} - \frac{π}{16} + \frac{πn}{2}

x = \frac{π}{2} - \frac{π}{16} + \frac{πn}{2}

x = \frac{π}{2} - \frac{π}{16} + \frac{πn}{2}

x = \frac{π}{2} - \frac{π}{16} + \frac{πn}{2}

x = \frac{π}{16} + \frac{πn}{2}, x = \frac{π}{2} - \frac{π}{16} + \frac{πn}{2}

cos (4 x) = - \frac{1}{2} : x = \frac{3 π}{16} + \frac{πn}{2}, x = - \frac{3 π}{16} + \frac{πn}{2}

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

使用反三角函数性质

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

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

的通解

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

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

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

解

4 x = arccos (- \frac{1}{2}) + 2 πn : x = \frac{3 π}{16} + \frac{πn}{2}

4 x = arccos (- \frac{1}{2}) + 2 πn

化简

arccos (- \frac{1}{2}) + 2 πn : \frac{3 π}{4} + 2 πn

arccos (- \frac{1}{2}) + 2 πn

使用以下普通恒等式:

arccos (- \frac{1}{2}) = \frac{3 π}{4}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{3 π}{4} + 2 πn

4 x = \frac{3 π}{4} + 2 πn

两边除以

4

4 x = \frac{3 π}{4} + 2 πn

两边除以

4

\frac{4 x}{4} = \frac{\frac{3 π}{4}}{4} + \frac{2 πn}{4}

化简

\frac{4 x}{4} = \frac{\frac{3 π}{4}}{4} + \frac{2 πn}{4}

化简

\frac{4 x}{4} : x

\frac{4 x}{4}

数字相除:

\frac{4}{4} = 1

= x

化简

\frac{\frac{3 π}{4}}{4} + \frac{2 πn}{4} : \frac{3 π}{16} + \frac{πn}{2}

\frac{\frac{3 π}{4}}{4} + \frac{2 πn}{4}

\frac{\frac{3 π}{4}}{4} = \frac{3 π}{16}

\frac{\frac{3 π}{4}}{4}

使用分式法则:

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

= \frac{3 π}{4 \cdot 4}

数字相乘:

4 \cdot 4 = 16

= \frac{3 π}{16}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

约分:

2

= \frac{πn}{2}

= \frac{3 π}{16} + \frac{πn}{2}

x = \frac{3 π}{16} + \frac{πn}{2}

x = \frac{3 π}{16} + \frac{πn}{2}

x = \frac{3 π}{16} + \frac{πn}{2}

解

4 x = - arccos (- \frac{1}{2}) + 2 πn : x = - \frac{3 π}{16} + \frac{πn}{2}

4 x = - arccos (- \frac{1}{2}) + 2 πn

化简

- arccos (- \frac{1}{2}) + 2 πn : - \frac{3 π}{4} + 2 πn

- arccos (- \frac{1}{2}) + 2 πn

使用以下普通恒等式:

arccos (- \frac{1}{2}) = \frac{3 π}{4}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= - \frac{3 π}{4} + 2 πn

4 x = - \frac{3 π}{4} + 2 πn

两边除以

4

4 x = - \frac{3 π}{4} + 2 πn

两边除以

4

\frac{4 x}{4} = - \frac{\frac{3 π}{4}}{4} + \frac{2 πn}{4}

化简

\frac{4 x}{4} = - \frac{\frac{3 π}{4}}{4} + \frac{2 πn}{4}

化简

\frac{4 x}{4} : x

\frac{4 x}{4}

数字相除:

\frac{4}{4} = 1

= x

化简

- \frac{\frac{3 π}{4}}{4} + \frac{2 πn}{4} : - \frac{3 π}{16} + \frac{πn}{2}

- \frac{\frac{3 π}{4}}{4} + \frac{2 πn}{4}

\frac{\frac{3 π}{4}}{4} = \frac{3 π}{16}

\frac{\frac{3 π}{4}}{4}

使用分式法则:

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

= \frac{3 π}{4 \cdot 4}

数字相乘:

4 \cdot 4 = 16

= \frac{3 π}{16}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

约分:

2

= \frac{πn}{2}

= - \frac{3 π}{16} + \frac{πn}{2}

x = - \frac{3 π}{16} + \frac{πn}{2}

x = - \frac{3 π}{16} + \frac{πn}{2}

x = - \frac{3 π}{16} + \frac{πn}{2}

x = \frac{3 π}{16} + \frac{πn}{2}, x = - \frac{3 π}{16} + \frac{πn}{2}

合并所有解

x = \frac{π}{16} + \frac{πn}{2}, x = \frac{π}{2} - \frac{π}{16} + \frac{πn}{2}, x = \frac{3 π}{16} + \frac{πn}{2}, x = - \frac{3 π}{16} + \frac{πn}{2}

作图

流行的例子

cos(θ)=-1/(sqrt(2))

cos (θ) = - \frac{1}{2}

3 sin (2 t) + 4 = 1

solvefor x,cos(x)=0

so l v e f or x, cos (x) = 0

cos(x+pi/6)-cos(x-pi/6)=1

cos (x + \frac{π}{6}) - cos (x - \frac{π}{6}) = 1

tan(2θ)+2sin(θ)=0

tan (2 θ) + 2 sin (θ) = 0