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受欢迎的三角函数 >

cos(x)+sin(x-1)=0

解答

cos (x) + sin (x - 1) = 0

解答

x = - \frac{π}{4} + \frac{1}{2} - 2 πn, x = \frac{1}{2} - \frac{5 π}{4} - 2 πn

+1

度数

x = - 16.35211 \dots^{\circ} - 36 0^{\circ} n, x = - 196.35211 \dots^{\circ} - 36 0^{\circ} n

求解步骤

cos (x) + sin (x - 1) = 0

使用三角恒等式改写

cos (x) + sin (- 1 + x)

利用以下特性:

sin (x) = cos (\frac{π}{2} - x)

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

- (- 1 + x) : 1 - x

- (- 1 + x)

打开括号

= - (- 1) - (x)

使用加减运算法则

- (- a) = a, - (a) = - a

= 1 - x

= cos (x) + cos (\frac{π}{2} + 1 - x)

使用和差化积恒等式:

cos (s) + cos (t) = 2 cos (\frac{s + t}{2}) cos (\frac{s - t}{2})

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

化简

2 cos (\frac{1 + \frac{π}{2} - x + x}{2}) cos (\frac{1 + \frac{π}{2} - x - x}{2}) : 2 cos (\frac{2 + π}{4}) cos (\frac{2 + π - 4 x}{4})

2 cos (\frac{1 + \frac{π}{2} - x + x}{2}) cos (\frac{1 + \frac{π}{2} - x - x}{2})

\frac{1 + \frac{π}{2} - x + x}{2} = \frac{2 + π}{4}

\frac{1 + \frac{π}{2} - x + x}{2}

同类项相加:

- x + x = 0

= \frac{1 + \frac{π}{2}}{2}

化简

1 + \frac{π}{2} : \frac{2 + π}{2}

1 + \frac{π}{2}

将项转换为分式:

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

= \frac{1 \cdot 2}{2} + \frac{π}{2}

因为分母相等，所以合并分式:

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

= \frac{1 \cdot 2 + π}{2}

数字相乘:

1 \cdot 2 = 2

= \frac{2 + π}{2}

= \frac{\frac{2 + π}{2}}{2}

使用分式法则:

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

= \frac{2 + π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{2 + π}{4}

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

\frac{1 + \frac{π}{2} - x - x}{2} = \frac{2 + π - 4 x}{4}

\frac{1 + \frac{π}{2} - x - x}{2}

同类项相加:

- x - x = - 2 x

= \frac{1 + \frac{π}{2} - 2 x}{2}

化简

1 + \frac{π}{2} - 2 x : \frac{2 + π - 4 x}{2}

1 + \frac{π}{2} - 2 x

将项转换为分式:

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

= \frac{1 \cdot 2}{2} + \frac{π}{2} - \frac{2 x \cdot 2}{2}

因为分母相等，所以合并分式:

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

= \frac{1 \cdot 2 + π - 2 x \cdot 2}{2}

1 \cdot 2 + π - 2 x \cdot 2 = 2 + π - 4 x

1 \cdot 2 + π - 2 x \cdot 2

数字相乘:

1 \cdot 2 = 2

= 2 + π - 2 \cdot 2 x

数字相乘:

2 \cdot 2 = 4

= 2 + π - 4 x

= \frac{2 + π - 4 x}{2}

= \frac{\frac{2 + π - 4 x}{2}}{2}

使用分式法则:

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

= \frac{2 + π - 4 x}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{2 + π - 4 x}{4}

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

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

2 cos (\frac{2 + π}{4}) cos (\frac{2 + π - 4 x}{4}) = 0

两边除以

2 cos (\frac{2 + π}{4})

2 cos (\frac{2 + π}{4}) cos (\frac{2 + π - 4 x}{4}) = 0

两边除以

2 cos (\frac{2 + π}{4})

\frac{2 cos ( \frac{2 + π}{4} ) cos ( \frac{2 + π - 4 x}{4} )}{2 cos ( \frac{2 + π}{4} )} = \frac{0}{2 cos ( \frac{2 + π}{4} )}

化简

cos (\frac{2 + π - 4 x}{4}) = 0

cos (\frac{2 + π - 4 x}{4}) = 0

cos (\frac{2 + π - 4 x}{4}) = 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}

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

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

解

\frac{2 + π - 4 x}{4} = \frac{π}{2} + 2 πn : x = - \frac{π}{4} + \frac{1}{2} - 2 πn

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

在两边乘以

4

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

在两边乘以

4

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

化简

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

化简

\frac{4 ( 2 + π - 4 x )}{4} : 2 + π - 4 x

\frac{4 ( 2 + π - 4 x )}{4}

数字相除:

\frac{4}{4} = 1

= 2 + π - 4 x

化简

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

4 \cdot \frac{π}{2} + 4 \cdot 2 πn

4 \cdot \frac{π}{2} = 2 π

4 \cdot \frac{π}{2}

分式相乘:

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

= \frac{π 4}{2}

数字相除:

\frac{4}{2} = 2

= 2 π

4 \cdot 2 πn = 8 πn

4 \cdot 2 πn

数字相乘:

4 \cdot 2 = 8

= 8 πn

= 2 π + 8 πn

2 + π - 4 x = 2 π + 8 πn

2 + π - 4 x = 2 π + 8 πn

2 + π - 4 x = 2 π + 8 πn

将

2

到右边

2 + π - 4 x = 2 π + 8 πn

两边减去

2

2 + π - 4 x - 2 = 2 π + 8 πn - 2

化简

π - 4 x = 2 π + 8 πn - 2

π - 4 x = 2 π + 8 πn - 2

将

π

到右边

π - 4 x = 2 π + 8 πn - 2

两边减去

π

π - 4 x - π = 2 π + 8 πn - 2 - π

化简

- 4 x = π + 8 πn - 2

- 4 x = π + 8 πn - 2

两边除以

- 4

- 4 x = π + 8 πn - 2

两边除以

- 4

\frac{- 4 x}{- 4} = \frac{π}{- 4} + \frac{8 πn}{- 4} - \frac{2}{- 4}

化简

\frac{- 4 x}{- 4} = \frac{π}{- 4} + \frac{8 πn}{- 4} - \frac{2}{- 4}

化简

\frac{- 4 x}{- 4} : x

\frac{- 4 x}{- 4}

使用分式法则:

\frac{- a}{- b} = \frac{a}{b}

= \frac{4 x}{4}

数字相除:

\frac{4}{4} = 1

= x

化简

\frac{π}{- 4} + \frac{8 πn}{- 4} - \frac{2}{- 4} : - \frac{π}{4} + \frac{1}{2} - 2 πn

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

对同类项分组

= \frac{π}{- 4} - \frac{2}{- 4} + \frac{8 πn}{- 4}

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{π}{4} - \frac{2}{- 4} + \frac{8 πn}{- 4}

\frac{2}{- 4} = - \frac{1}{2}

\frac{2}{- 4}

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{2}{4}

约分:

2

= - \frac{1}{2}

\frac{8 πn}{- 4} = - 2 πn

\frac{8 πn}{- 4}

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{8 πn}{4}

数字相除:

\frac{8}{4} = 2

= - 2 πn

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

使用法则

- (- a) = a

= - \frac{π}{4} + \frac{1}{2} - 2 πn

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

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

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

解

\frac{2 + π - 4 x}{4} = \frac{3 π}{2} + 2 πn : x = \frac{1}{2} - \frac{5 π}{4} - 2 πn

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

在两边乘以

4

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

在两边乘以

4

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

化简

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

化简

\frac{4 ( 2 + π - 4 x )}{4} : 2 + π - 4 x

\frac{4 ( 2 + π - 4 x )}{4}

数字相除:

\frac{4}{4} = 1

= 2 + π - 4 x

化简

4 \cdot \frac{3 π}{2} + 4 \cdot 2 πn : 6 π + 8 πn

4 \cdot \frac{3 π}{2} + 4 \cdot 2 πn

4 \cdot \frac{3 π}{2} = 6 π

4 \cdot \frac{3 π}{2}

分式相乘:

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

= \frac{3 π 4}{2}

数字相乘:

3 \cdot 4 = 12

= \frac{12 π}{2}

数字相除:

\frac{12}{2} = 6

= 6 π

4 \cdot 2 πn = 8 πn

4 \cdot 2 πn

数字相乘:

4 \cdot 2 = 8

= 8 πn

= 6 π + 8 πn

2 + π - 4 x = 6 π + 8 πn

2 + π - 4 x = 6 π + 8 πn

2 + π - 4 x = 6 π + 8 πn

将

2

到右边

2 + π - 4 x = 6 π + 8 πn

两边减去

2

2 + π - 4 x - 2 = 6 π + 8 πn - 2

化简

π - 4 x = 6 π + 8 πn - 2

π - 4 x = 6 π + 8 πn - 2

将

π

到右边

π - 4 x = 6 π + 8 πn - 2

两边减去

π

π - 4 x - π = 6 π + 8 πn - 2 - π

化简

- 4 x = 5 π + 8 πn - 2

- 4 x = 5 π + 8 πn - 2

两边除以

- 4

- 4 x = 5 π + 8 πn - 2

两边除以

- 4

\frac{- 4 x}{- 4} = \frac{5 π}{- 4} + \frac{8 πn}{- 4} - \frac{2}{- 4}

化简

\frac{- 4 x}{- 4} = \frac{5 π}{- 4} + \frac{8 πn}{- 4} - \frac{2}{- 4}

化简

\frac{- 4 x}{- 4} : x

\frac{- 4 x}{- 4}

使用分式法则:

\frac{- a}{- b} = \frac{a}{b}

= \frac{4 x}{4}

数字相除:

\frac{4}{4} = 1

= x

化简

\frac{5 π}{- 4} + \frac{8 πn}{- 4} - \frac{2}{- 4} : \frac{1}{2} - \frac{5 π}{4} - 2 πn

\frac{5 π}{- 4} + \frac{8 πn}{- 4} - \frac{2}{- 4}

对同类项分组

= - \frac{2}{- 4} + \frac{5 π}{- 4} + \frac{8 πn}{- 4}

\frac{2}{- 4} = - \frac{1}{2}

\frac{2}{- 4}

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{2}{4}

约分:

2

= - \frac{1}{2}

\frac{5 π}{- 4} = - \frac{5 π}{4}

\frac{5 π}{- 4}

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{5 π}{4}

= - (- \frac{1}{2}) - \frac{5 π}{4} + \frac{8 πn}{- 4}

使用法则

- (- a) = a

= \frac{1}{2} - \frac{5 π}{4} + \frac{8 πn}{- 4}

\frac{8 πn}{- 4} = - 2 πn

\frac{8 πn}{- 4}

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{8 πn}{4}

数字相除:

\frac{8}{4} = 2

= - 2 πn

= \frac{1}{2} - \frac{5 π}{4} - 2 πn

x = \frac{1}{2} - \frac{5 π}{4} - 2 πn

x = \frac{1}{2} - \frac{5 π}{4} - 2 πn

x = \frac{1}{2} - \frac{5 π}{4} - 2 πn

x = - \frac{π}{4} + \frac{1}{2} - 2 πn, x = \frac{1}{2} - \frac{5 π}{4} - 2 πn

作图

流行的例子

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