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 >

1.5=1+0.6cos((pit)/2)

解答

1.5 = 1 + 0.6 cos (\frac{π t}{2})

解答

t = \frac{2 \cdot 0.58568 \dots}{π} + 4 n, t = 4 - \frac{2 \cdot 0.58568 \dots}{π} + 4 n

+1

度数

t = 21.36324 \dots^{\circ} + 229.18311 \dots^{\circ} n, t = 207.81987 \dots^{\circ} + 229.18311 \dots^{\circ} n

求解步骤

1.5 = 1 + 0.6 cos (\frac{π t}{2})

交换两边

1 + 0.6 cos (\frac{π t}{2}) = 1.5

在两边乘以

10

1 + 0.6 cos (\frac{π t}{2}) = 1.5

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere is one digit to the right of the decimal point, therefore multiply by

10

1 \cdot 10 + 0.6 cos (\frac{π t}{2}) \cdot 10 = 1.5 \cdot 10

整理后得

10 + 6 cos (\frac{π t}{2}) = 15

10 + 6 cos (\frac{π t}{2}) = 15

将

10

到右边

10 + 6 cos (\frac{π t}{2}) = 15

两边减去

10

10 + 6 cos (\frac{π t}{2}) - 10 = 15 - 10

化简

6 cos (\frac{π t}{2}) = 5

6 cos (\frac{π t}{2}) = 5

两边除以

6

6 cos (\frac{π t}{2}) = 5

两边除以

6

\frac{6 cos ( \frac{π t}{2} )}{6} = \frac{5}{6}

化简

cos (\frac{π t}{2}) = \frac{5}{6}

cos (\frac{π t}{2}) = \frac{5}{6}

使用反三角函数性质

cos (\frac{π t}{2}) = \frac{5}{6}

cos (\frac{π t}{2}) = \frac{5}{6}

的通解

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

\frac{π t}{2} = arccos (\frac{5}{6}) + 2 πn, \frac{π t}{2} = 2 π - arccos (\frac{5}{6}) + 2 πn

\frac{π t}{2} = arccos (\frac{5}{6}) + 2 πn, \frac{π t}{2} = 2 π - arccos (\frac{5}{6}) + 2 πn

解

\frac{π t}{2} = arccos (\frac{5}{6}) + 2 πn : t = \frac{2 arccos ( \frac{5}{6} )}{π} + 4 n

\frac{π t}{2} = arccos (\frac{5}{6}) + 2 πn

在两边乘以

2

\frac{π t}{2} = arccos (\frac{5}{6}) + 2 πn

在两边乘以

2

\frac{2 π t}{2} = 2 arccos (\frac{5}{6}) + 2 \cdot 2 πn

化简

\frac{2 π t}{2} = 2 arccos (\frac{5}{6}) + 2 \cdot 2 πn

化简

\frac{2 π t}{2} : π t

\frac{2 π t}{2}

数字相除:

\frac{2}{2} = 1

= π t

化简

2 arccos (\frac{5}{6}) + 2 \cdot 2 πn : 2 arccos (\frac{5}{6}) + 4 πn

2 arccos (\frac{5}{6}) + 2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 2 arccos (\frac{5}{6}) + 4 πn

π t = 2 arccos (\frac{5}{6}) + 4 πn

π t = 2 arccos (\frac{5}{6}) + 4 πn

π t = 2 arccos (\frac{5}{6}) + 4 πn

两边除以

π

π t = 2 arccos (\frac{5}{6}) + 4 πn

两边除以

π

\frac{π t}{π} = \frac{2 arccos ( \frac{5}{6} )}{π} + \frac{4 πn}{π}

化简

t = \frac{2 arccos ( \frac{5}{6} )}{π} + 4 n

t = \frac{2 arccos ( \frac{5}{6} )}{π} + 4 n

解

\frac{π t}{2} = 2 π - arccos (\frac{5}{6}) + 2 πn : t = 4 - \frac{2 arccos ( \frac{5}{6} )}{π} + 4 n

\frac{π t}{2} = 2 π - arccos (\frac{5}{6}) + 2 πn

在两边乘以

2

\frac{π t}{2} = 2 π - arccos (\frac{5}{6}) + 2 πn

在两边乘以

2

\frac{2 π t}{2} = 2 \cdot 2 π - 2 arccos (\frac{5}{6}) + 2 \cdot 2 πn

化简

\frac{2 π t}{2} = 2 \cdot 2 π - 2 arccos (\frac{5}{6}) + 2 \cdot 2 πn

化简

\frac{2 π t}{2} : π t

\frac{2 π t}{2}

数字相除:

\frac{2}{2} = 1

= π t

化简

2 \cdot 2 π - 2 arccos (\frac{5}{6}) + 2 \cdot 2 πn : 4 π - 2 arccos (\frac{5}{6}) + 4 πn

2 \cdot 2 π - 2 arccos (\frac{5}{6}) + 2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 π - 2 arccos (\frac{5}{6}) + 4 πn

π t = 4 π - 2 arccos (\frac{5}{6}) + 4 πn

π t = 4 π - 2 arccos (\frac{5}{6}) + 4 πn

π t = 4 π - 2 arccos (\frac{5}{6}) + 4 πn

两边除以

π

π t = 4 π - 2 arccos (\frac{5}{6}) + 4 πn

两边除以

π

\frac{π t}{π} = \frac{4 π}{π} - \frac{2 arccos ( \frac{5}{6} )}{π} + \frac{4 πn}{π}

化简

\frac{π t}{π} = \frac{4 π}{π} - \frac{2 arccos ( \frac{5}{6} )}{π} + \frac{4 πn}{π}

化简

\frac{π t}{π} : t

\frac{π t}{π}

约分:

π

= t

化简

\frac{4 π}{π} - \frac{2 arccos ( \frac{5}{6} )}{π} + \frac{4 πn}{π} : 4 - \frac{2 arccos ( \frac{5}{6} )}{π} + 4 n

\frac{4 π}{π} - \frac{2 arccos ( \frac{5}{6} )}{π} + \frac{4 πn}{π}

消掉

\frac{4 π}{π} : 4

\frac{4 π}{π}

约分:

π

= 4

= 4 - \frac{2 arccos ( \frac{5}{6} )}{π} + \frac{4 πn}{π}

消掉

\frac{4 πn}{π} : 4 n

\frac{4 πn}{π}

约分:

π

= 4 n

= 4 - \frac{2 arccos ( \frac{5}{6} )}{π} + 4 n

t = 4 - \frac{2 arccos ( \frac{5}{6} )}{π} + 4 n

t = 4 - \frac{2 arccos ( \frac{5}{6} )}{π} + 4 n

t = 4 - \frac{2 arccos ( \frac{5}{6} )}{π} + 4 n

t = \frac{2 arccos ( \frac{5}{6} )}{π} + 4 n, t = 4 - \frac{2 arccos ( \frac{5}{6} )}{π} + 4 n

以小数形式表示解

t = \frac{2 \cdot 0.58568 \dots}{π} + 4 n, t = 4 - \frac{2 \cdot 0.58568 \dots}{π} + 4 n

作图

流行的例子

tan(x)csc(x)=cos(2)

tan (x) csc (x) = cos (2)

solvefor x,z=e^ysin(x)y

so l v e f or x, z = e^{y} sin (x) y

tan(θ)sec(θ)=cot(θ)csc(θ)

tan (θ) sec (θ) = cot (θ) csc (θ)

0 = 4 + 3 cos (x)

cos^2(x)-cos(x)=0.75

cos^{2} (x) - cos (x) = 0.75