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(4θ)=0

解答

2 cos (4 θ) = 0

解答

θ = \frac{π}{8} + \frac{πn}{2}, θ = \frac{3 π}{8} + \frac{πn}{2}

+1

度数

θ = 22. 5^{\circ} + 9 0^{\circ} n, θ = 67. 5^{\circ} + 9 0^{\circ} n

求解步骤

2 cos (4 θ) = 0

两边除以

2

2 cos (4 θ) = 0

两边除以

2

\frac{2 cos ( 4 θ )}{2} = \frac{0}{2}

化简

cos (4 θ) = 0

cos (4 θ) = 0

cos (4 θ) = 0

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}

4 θ = \frac{π}{2} + 2 πn, 4 θ = \frac{3 π}{2} + 2 πn

4 θ = \frac{π}{2} + 2 πn, 4 θ = \frac{3 π}{2} + 2 πn

解

4 θ = \frac{π}{2} + 2 πn : θ = \frac{π}{8} + \frac{πn}{2}

4 θ = \frac{π}{2} + 2 πn

两边除以

4

4 θ = \frac{π}{2} + 2 πn

两边除以

4

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

化简

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

化简

\frac{4 θ}{4} : θ

\frac{4 θ}{4}

数字相除:

\frac{4}{4} = 1

= θ

化简

\frac{\frac{π}{2}}{4} + \frac{2 πn}{4} : \frac{π}{8} + \frac{πn}{2}

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

\frac{\frac{π}{2}}{4} = \frac{π}{8}

\frac{\frac{π}{2}}{4}

使用分式法则:

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

= \frac{π}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{π}{8}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

约分:

2

= \frac{πn}{2}

= \frac{π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}

解

4 θ = \frac{3 π}{2} + 2 πn : θ = \frac{3 π}{8} + \frac{πn}{2}

4 θ = \frac{3 π}{2} + 2 πn

两边除以

4

4 θ = \frac{3 π}{2} + 2 πn

两边除以

4

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

化简

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

化简

\frac{4 θ}{4} : θ

\frac{4 θ}{4}

数字相除:

\frac{4}{4} = 1

= θ

化简

\frac{\frac{3 π}{2}}{4} + \frac{2 πn}{4} : \frac{3 π}{8} + \frac{πn}{2}

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

\frac{\frac{3 π}{2}}{4} = \frac{3 π}{8}

\frac{\frac{3 π}{2}}{4}

使用分式法则:

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

= \frac{3 π}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{3 π}{8}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

约分:

2

= \frac{πn}{2}

= \frac{3 π}{8} + \frac{πn}{2}

θ = \frac{3 π}{8} + \frac{πn}{2}

θ = \frac{3 π}{8} + \frac{πn}{2}

θ = \frac{3 π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}, θ = \frac{3 π}{8} + \frac{πn}{2}

作图

流行的例子

1032/3145 sin(4x)+2196/3145 cos(4x)=0

\frac{1032}{3145} sin (4 x) + \frac{2196}{3145} cos (4 x) = 0

6-2sin(θ)-8sin^2(θ)=0

6 - 2 sin (θ) - 8 sin^{2} (θ) = 0

cos(2θ)=cos(3θ)

cos (2 θ) = cos (3 θ)

solvefor θ,2sin(θ)=1.5

so l v e f or θ, 2 sin (θ) = 1.5

cos (θ) = - 2109