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 >

3+cos(4θ)=2-cos(4θ)

解答

3 + cos (4 θ) = 2 - cos (4 θ)

解答

θ = \frac{π}{6} + \frac{πn}{2}, θ = \frac{π}{3} + \frac{πn}{2}

+1

度数

θ = 3 0^{\circ} + 9 0^{\circ} n, θ = 6 0^{\circ} + 9 0^{\circ} n

求解步骤

3 + cos (4 θ) = 2 - cos (4 θ)

用替代法求解

3 + cos (4 θ) = 2 - cos (4 θ)

令

: cos (4 θ) = u

3 + u = 2 - u

3 + u = 2 - u : u = - \frac{1}{2}

3 + u = 2 - u

将

3

到右边

3 + u = 2 - u

两边减去

3

3 + u - 3 = 2 - u - 3

化简

u = - u - 1

u = - u - 1

将

u

para o lado esquerdo

u = - u - 1

两边加上

u

u + u = - u - 1 + u

化简

2 u = - 1

2 u = - 1

两边除以

2

2 u = - 1

两边除以

2

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

化简

u = - \frac{1}{2}

u = - \frac{1}{2}

u = cos (4 θ)

代回

cos (4 θ) = - \frac{1}{2}

cos (4 θ) = - \frac{1}{2}

cos (4 θ) = - \frac{1}{2} : θ = \frac{π}{6} + \frac{πn}{2}, θ = \frac{π}{3} + \frac{πn}{2}

cos (4 θ) = - \frac{1}{2}

cos (4 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}

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

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

解

4 θ = \frac{2 π}{3} + 2 πn : θ = \frac{π}{6} + \frac{πn}{2}

4 θ = \frac{2 π}{3} + 2 πn

两边除以

4

4 θ = \frac{2 π}{3} + 2 πn

两边除以

4

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

化简

\frac{4 θ}{4} = \frac{\frac{2 π}{3}}{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{2 π}{3}}{4} + \frac{2 πn}{4} : \frac{π}{6} + \frac{πn}{2}

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

\frac{\frac{2 π}{3}}{4} = \frac{π}{6}

\frac{\frac{2 π}{3}}{4}

使用分式法则:

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

= \frac{2 π}{3 \cdot 4}

数字相乘:

3 \cdot 4 = 12

= \frac{2 π}{12}

约分:

2

= \frac{π}{6}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

约分:

2

= \frac{πn}{2}

= \frac{π}{6} + \frac{πn}{2}

θ = \frac{π}{6} + \frac{πn}{2}

θ = \frac{π}{6} + \frac{πn}{2}

θ = \frac{π}{6} + \frac{πn}{2}

解

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

4 θ = \frac{4 π}{3} + 2 πn

两边除以

4

4 θ = \frac{4 π}{3} + 2 πn

两边除以

4

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

化简

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

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

\frac{\frac{4 π}{3}}{4} = \frac{π}{3}

\frac{\frac{4 π}{3}}{4}

使用分式法则:

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

= \frac{4 π}{3 \cdot 4}

数字相乘:

3 \cdot 4 = 12

= \frac{4 π}{12}

约分:

4

= \frac{π}{3}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

约分:

2

= \frac{πn}{2}

= \frac{π}{3} + \frac{πn}{2}

θ = \frac{π}{3} + \frac{πn}{2}

θ = \frac{π}{3} + \frac{πn}{2}

θ = \frac{π}{3} + \frac{πn}{2}

θ = \frac{π}{6} + \frac{πn}{2}, θ = \frac{π}{3} + \frac{πn}{2}

合并所有解

θ = \frac{π}{6} + \frac{πn}{2}, θ = \frac{π}{3} + \frac{πn}{2}

作图

流行的例子

5sec(θ)=4sec(θ)+sqrt(2)

5 sec (θ) = 4 sec (θ) + 2

4 sec (π x) + 6 = - 2

solvefor x,y-1=-4+3sin(4x-pi/4)

so l v e f or x, y - 1 = - 4 + 3 sin (4 x - \frac{π}{4})

64=81+25-90(cos(C))

64 = 81 + 25 - 90 (cos (C))

sin^2(a-pi/8)= 2/3

sin^{2} (a - \frac{π}{8}) = \frac{2}{3}